Let $g(k)$ be the greatest odd divisor of $k$ show that $ 0< sum_{k=1}^n frac {g(k)}{k} - frac {2n}{3} lt...












8












$begingroup$



Prove that for all positive intergers $n$,



$$ 0< sum_{k=1}^n frac {g(k)}{k} - frac {2n}{3} < frac {2}{3}$$



Where $g(k)$ denotes the greatest odd divisor of $k$.




Here's my try:



All numbers $k$ can be written as $k= 2^ts$, for nonnegative $t$ and odd $s$, therefore if $k= 2^ts$, then $frac {g(k)}{k} = frac {1}{2^k}$ i.e $frac {g(k)}{k}$ is equal to $1$ divided by the highest power of $2$ dividing $k$. I first thought of proving the inequality for $k= 2^n (n>1)$. Let $Q= sum_{k=1}^{2^n} frac {g(k)}{k}$, then:



$$Q = frac {1}{2}q_1 +frac {1}{2^2}q_2 + cdots + frac {1}{2^{n -1}} q_{2^{n -1}}+ frac {1}{2^n} q_{2^n}+ 2^{n-1} $$.



$q_i$ is the number of times $frac {1}{2^i}$ is added in the summation. It's easy to notice that $q_{2^n} =1$. $q_i$ for $0< i < 2^n$ would be equal to $2^{n-1-i}$ (comes from this $(2^i)(2(2^{n-1-i})-1)$). Then:



$$Q= 2^{n-1}+ frac {1}{2^n} + sum_{i=1}^{n-1} frac{1}{2^i} cdot 2^{n-1-i} $$



$$Q = 2^{n-1}+ frac {1}{2^n} + 2^{n-1} sum_{i=1}^{n-1} (frac{1}{4})^i $$



EDITED



With some algebra we get that:



$$Q - frac {2}{3} cdot 2^n= - frac {4}{3} cdot 2^{n-1} + 2^{n-1}+ frac {1}{2^n} + 2^{n-1} cdot frac {4^{n-1}-1}{4^{n-1}} cdot frac {1}{3} < frac {1}{2^n} < frac {2}{3} $$



Also:



$$ Q - frac {2}{3} cdot 2^n = frac {1}{2^n} - frac {2^{n-1}}{4^{n-1}} cdot frac {1}{3}= frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{3}> frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{2}= 0$$



Then I would get:



$$ 0< Q - frac {2}{3} cdot 2^n < frac {2}{3}$$



Well, I thought proving the inequality for $k=2^n$ because I had some idea about how many times the powers of $2$ appeared. I then thought that I could go backwards with induction but I'm stuck. I would like to see some other approaches but I would also like to know if it's possible to solve the problem from the point where I am.



I'll appreciate any hints or help, thanks in advance.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
    $endgroup$
    – Marco
    Dec 16 '18 at 2:03










  • $begingroup$
    Your summation is missing a $q_0$ term.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:15










  • $begingroup$
    @Marco you are right, thank you.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21










  • $begingroup$
    @JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21
















8












$begingroup$



Prove that for all positive intergers $n$,



$$ 0< sum_{k=1}^n frac {g(k)}{k} - frac {2n}{3} < frac {2}{3}$$



Where $g(k)$ denotes the greatest odd divisor of $k$.




Here's my try:



All numbers $k$ can be written as $k= 2^ts$, for nonnegative $t$ and odd $s$, therefore if $k= 2^ts$, then $frac {g(k)}{k} = frac {1}{2^k}$ i.e $frac {g(k)}{k}$ is equal to $1$ divided by the highest power of $2$ dividing $k$. I first thought of proving the inequality for $k= 2^n (n>1)$. Let $Q= sum_{k=1}^{2^n} frac {g(k)}{k}$, then:



$$Q = frac {1}{2}q_1 +frac {1}{2^2}q_2 + cdots + frac {1}{2^{n -1}} q_{2^{n -1}}+ frac {1}{2^n} q_{2^n}+ 2^{n-1} $$.



$q_i$ is the number of times $frac {1}{2^i}$ is added in the summation. It's easy to notice that $q_{2^n} =1$. $q_i$ for $0< i < 2^n$ would be equal to $2^{n-1-i}$ (comes from this $(2^i)(2(2^{n-1-i})-1)$). Then:



$$Q= 2^{n-1}+ frac {1}{2^n} + sum_{i=1}^{n-1} frac{1}{2^i} cdot 2^{n-1-i} $$



$$Q = 2^{n-1}+ frac {1}{2^n} + 2^{n-1} sum_{i=1}^{n-1} (frac{1}{4})^i $$



EDITED



With some algebra we get that:



$$Q - frac {2}{3} cdot 2^n= - frac {4}{3} cdot 2^{n-1} + 2^{n-1}+ frac {1}{2^n} + 2^{n-1} cdot frac {4^{n-1}-1}{4^{n-1}} cdot frac {1}{3} < frac {1}{2^n} < frac {2}{3} $$



Also:



$$ Q - frac {2}{3} cdot 2^n = frac {1}{2^n} - frac {2^{n-1}}{4^{n-1}} cdot frac {1}{3}= frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{3}> frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{2}= 0$$



Then I would get:



$$ 0< Q - frac {2}{3} cdot 2^n < frac {2}{3}$$



Well, I thought proving the inequality for $k=2^n$ because I had some idea about how many times the powers of $2$ appeared. I then thought that I could go backwards with induction but I'm stuck. I would like to see some other approaches but I would also like to know if it's possible to solve the problem from the point where I am.



I'll appreciate any hints or help, thanks in advance.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
    $endgroup$
    – Marco
    Dec 16 '18 at 2:03










  • $begingroup$
    Your summation is missing a $q_0$ term.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:15










  • $begingroup$
    @Marco you are right, thank you.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21










  • $begingroup$
    @JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21














8












8








8


4



$begingroup$



Prove that for all positive intergers $n$,



$$ 0< sum_{k=1}^n frac {g(k)}{k} - frac {2n}{3} < frac {2}{3}$$



Where $g(k)$ denotes the greatest odd divisor of $k$.




Here's my try:



All numbers $k$ can be written as $k= 2^ts$, for nonnegative $t$ and odd $s$, therefore if $k= 2^ts$, then $frac {g(k)}{k} = frac {1}{2^k}$ i.e $frac {g(k)}{k}$ is equal to $1$ divided by the highest power of $2$ dividing $k$. I first thought of proving the inequality for $k= 2^n (n>1)$. Let $Q= sum_{k=1}^{2^n} frac {g(k)}{k}$, then:



$$Q = frac {1}{2}q_1 +frac {1}{2^2}q_2 + cdots + frac {1}{2^{n -1}} q_{2^{n -1}}+ frac {1}{2^n} q_{2^n}+ 2^{n-1} $$.



$q_i$ is the number of times $frac {1}{2^i}$ is added in the summation. It's easy to notice that $q_{2^n} =1$. $q_i$ for $0< i < 2^n$ would be equal to $2^{n-1-i}$ (comes from this $(2^i)(2(2^{n-1-i})-1)$). Then:



$$Q= 2^{n-1}+ frac {1}{2^n} + sum_{i=1}^{n-1} frac{1}{2^i} cdot 2^{n-1-i} $$



$$Q = 2^{n-1}+ frac {1}{2^n} + 2^{n-1} sum_{i=1}^{n-1} (frac{1}{4})^i $$



EDITED



With some algebra we get that:



$$Q - frac {2}{3} cdot 2^n= - frac {4}{3} cdot 2^{n-1} + 2^{n-1}+ frac {1}{2^n} + 2^{n-1} cdot frac {4^{n-1}-1}{4^{n-1}} cdot frac {1}{3} < frac {1}{2^n} < frac {2}{3} $$



Also:



$$ Q - frac {2}{3} cdot 2^n = frac {1}{2^n} - frac {2^{n-1}}{4^{n-1}} cdot frac {1}{3}= frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{3}> frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{2}= 0$$



Then I would get:



$$ 0< Q - frac {2}{3} cdot 2^n < frac {2}{3}$$



Well, I thought proving the inequality for $k=2^n$ because I had some idea about how many times the powers of $2$ appeared. I then thought that I could go backwards with induction but I'm stuck. I would like to see some other approaches but I would also like to know if it's possible to solve the problem from the point where I am.



I'll appreciate any hints or help, thanks in advance.










share|cite|improve this question











$endgroup$





Prove that for all positive intergers $n$,



$$ 0< sum_{k=1}^n frac {g(k)}{k} - frac {2n}{3} < frac {2}{3}$$



Where $g(k)$ denotes the greatest odd divisor of $k$.




Here's my try:



All numbers $k$ can be written as $k= 2^ts$, for nonnegative $t$ and odd $s$, therefore if $k= 2^ts$, then $frac {g(k)}{k} = frac {1}{2^k}$ i.e $frac {g(k)}{k}$ is equal to $1$ divided by the highest power of $2$ dividing $k$. I first thought of proving the inequality for $k= 2^n (n>1)$. Let $Q= sum_{k=1}^{2^n} frac {g(k)}{k}$, then:



$$Q = frac {1}{2}q_1 +frac {1}{2^2}q_2 + cdots + frac {1}{2^{n -1}} q_{2^{n -1}}+ frac {1}{2^n} q_{2^n}+ 2^{n-1} $$.



$q_i$ is the number of times $frac {1}{2^i}$ is added in the summation. It's easy to notice that $q_{2^n} =1$. $q_i$ for $0< i < 2^n$ would be equal to $2^{n-1-i}$ (comes from this $(2^i)(2(2^{n-1-i})-1)$). Then:



$$Q= 2^{n-1}+ frac {1}{2^n} + sum_{i=1}^{n-1} frac{1}{2^i} cdot 2^{n-1-i} $$



$$Q = 2^{n-1}+ frac {1}{2^n} + 2^{n-1} sum_{i=1}^{n-1} (frac{1}{4})^i $$



EDITED



With some algebra we get that:



$$Q - frac {2}{3} cdot 2^n= - frac {4}{3} cdot 2^{n-1} + 2^{n-1}+ frac {1}{2^n} + 2^{n-1} cdot frac {4^{n-1}-1}{4^{n-1}} cdot frac {1}{3} < frac {1}{2^n} < frac {2}{3} $$



Also:



$$ Q - frac {2}{3} cdot 2^n = frac {1}{2^n} - frac {2^{n-1}}{4^{n-1}} cdot frac {1}{3}= frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{3}> frac {1}{2^n} - frac {1}{2^{n-1}} cdot frac {1}{2}= 0$$



Then I would get:



$$ 0< Q - frac {2}{3} cdot 2^n < frac {2}{3}$$



Well, I thought proving the inequality for $k=2^n$ because I had some idea about how many times the powers of $2$ appeared. I then thought that I could go backwards with induction but I'm stuck. I would like to see some other approaches but I would also like to know if it's possible to solve the problem from the point where I am.



I'll appreciate any hints or help, thanks in advance.







sequences-and-series number-theory inequality divisibility analytic-number-theory






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edited Dec 18 '18 at 9:45









Batominovski

33.1k33293




33.1k33293










asked Dec 15 '18 at 5:19









VmimiVmimi

372212




372212












  • $begingroup$
    Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
    $endgroup$
    – Marco
    Dec 16 '18 at 2:03










  • $begingroup$
    Your summation is missing a $q_0$ term.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:15










  • $begingroup$
    @Marco you are right, thank you.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21










  • $begingroup$
    @JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21


















  • $begingroup$
    Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
    $endgroup$
    – Marco
    Dec 16 '18 at 2:03










  • $begingroup$
    Your summation is missing a $q_0$ term.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:15










  • $begingroup$
    @Marco you are right, thank you.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21










  • $begingroup$
    @JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
    $endgroup$
    – Vmimi
    Dec 17 '18 at 4:21
















$begingroup$
Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
$endgroup$
– Marco
Dec 16 '18 at 2:03




$begingroup$
Check the summation of $(1/4)^i$ part again. The 3 should come out in the denominator.
$endgroup$
– Marco
Dec 16 '18 at 2:03












$begingroup$
Your summation is missing a $q_0$ term.
$endgroup$
– JimmyK4542
Dec 16 '18 at 2:15




$begingroup$
Your summation is missing a $q_0$ term.
$endgroup$
– JimmyK4542
Dec 16 '18 at 2:15












$begingroup$
@Marco you are right, thank you.
$endgroup$
– Vmimi
Dec 17 '18 at 4:21




$begingroup$
@Marco you are right, thank you.
$endgroup$
– Vmimi
Dec 17 '18 at 4:21












$begingroup$
@JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
$endgroup$
– Vmimi
Dec 17 '18 at 4:21




$begingroup$
@JimmyK4542 Well, I forgot to say that but I did include it, it is in the part where I add $2^{n−1}$ since half of the numbers are odd.
$endgroup$
– Vmimi
Dec 17 '18 at 4:21










2 Answers
2






active

oldest

votes


















2












$begingroup$

Since $2^jmid k$ for each $jle v_2(k)$ and $sumlimits_{j=1}^n2^{-j}=1-2^{-n}$, we have
$$
2^{-v_2(k)}=1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}tag1
$$

where $[dots]$ are Iverson brackets. Thus,
$$
begin{align}
sum_{k=1}^nfrac{g(k)}k
&=sum_{k=1}^n2^{-v_2(k)}\
&=sum_{k=1}^nleft(1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}right)\
&=n-sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}tag2
end{align}
$$

Furthermore, since $leftlfloorfrac n{2^j}rightrfloorlefrac n{2^j}$ with equality iff $nequiv0pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&lesum_{j=1}^inftyfrac n{2^j}frac1{2^j}\
&=frac n3tag3
end{align}
$$

and $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$ with equality iff $nequiv-1pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&gesum_{j=1}^inftyfrac{n-(2^j-1)}{2^j}frac1{2^j}\
&=frac n3-frac23tag4
end{align}
$$

Equality in $(3)$ can only occur if $nequiv0pmod{2^j}$ for all $j$ ($implies n=0$) and equality in $(4)$ can only occur if $nequiv-1pmod{2^j}$ for all $j$ ($implies n=-1$). Therefore, for $nge1$,
$$
frac{2n}3ltsum_{k=1}^nfrac{g(k)}kltfrac{2n}3+frac23tag5
$$






share|cite|improve this answer











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  • $begingroup$
    I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:49










  • $begingroup$
    And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:56










  • $begingroup$
    For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
    $endgroup$
    – robjohn
    Dec 19 '18 at 21:40





















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$begingroup$

Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $dfrac{g(k)}{k} = 2^{-v(k)}$.



So, let $S_n = displaystylesum_{k = 1}^{n}dfrac{g(k)}{k} = sum_{k = 1}^{n}2^{-v(k)}$.



Then, we have:



begin{align*}
S_{2n} &= displaystylesum_{k = 1}^{2n}2^{-v(k)}
\
&= sum_{k = 1}^{n}2^{-v(2k-1)} + sum_{k = 1}^{n}2^{-v(2k)}
\
&= sum_{k = 1}^{n}2^{-0} + sum_{k = 1}^{n}2^{-(1+v(k))}
\
&= sum_{k = 1}^{n}1 + dfrac{1}{2}sum_{k = 1}^{n}2^{-v(k)}
\
&= n + dfrac{1}{2}S_n
end{align*}



Also, $S_{2n+1} = displaystylesum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.



We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > dfrac{2}{3}$ and $S_1 < dfrac{4}{3}$.



Now, suppose that for some integer $n$, we have $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$.



Then, we have:



$$S_{2n} = dfrac{1}{2}S_n + n > dfrac{1}{2}left(dfrac{2n}{3}right) + n = dfrac{4n}{3} = dfrac{2 cdot 2n}{3}$$



$$S_{2n} = dfrac{1}{2}S_n + n < dfrac{1}{2}left(dfrac{2n}{3}+dfrac{2}{3}right) + n = dfrac{4n}{3} + dfrac{1}{3} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$$



$$S_{2n+1} = S_{2n}+1 > left(dfrac{4n}{3}right)+1 > dfrac{4n}{3}+dfrac{2}{3} = dfrac{2(2n+1)}{3}$$



$$S_{2n+1} = S_{2n}+1 < left(dfrac{4n}{3}+dfrac{1}{3}right)+1 = dfrac{2(2n+1)}{3} + dfrac{2}{3}.$$



Hence, $dfrac{2 cdot 2n}{3} < S_{2n} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$ and $dfrac{2(2n+1)}{3} < S_{2n+1} < dfrac{2(2n+1)}{3} + dfrac{2}{3}$.



So by induction, we have that $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$ for all $n$, as desired.






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$endgroup$









  • 1




    $begingroup$
    So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 2:35








  • 1




    $begingroup$
    If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:46








  • 1




    $begingroup$
    So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 9:52






  • 1




    $begingroup$
    If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 10:43








  • 1




    $begingroup$
    I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 11:10













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2 Answers
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2 Answers
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2












$begingroup$

Since $2^jmid k$ for each $jle v_2(k)$ and $sumlimits_{j=1}^n2^{-j}=1-2^{-n}$, we have
$$
2^{-v_2(k)}=1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}tag1
$$

where $[dots]$ are Iverson brackets. Thus,
$$
begin{align}
sum_{k=1}^nfrac{g(k)}k
&=sum_{k=1}^n2^{-v_2(k)}\
&=sum_{k=1}^nleft(1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}right)\
&=n-sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}tag2
end{align}
$$

Furthermore, since $leftlfloorfrac n{2^j}rightrfloorlefrac n{2^j}$ with equality iff $nequiv0pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&lesum_{j=1}^inftyfrac n{2^j}frac1{2^j}\
&=frac n3tag3
end{align}
$$

and $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$ with equality iff $nequiv-1pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&gesum_{j=1}^inftyfrac{n-(2^j-1)}{2^j}frac1{2^j}\
&=frac n3-frac23tag4
end{align}
$$

Equality in $(3)$ can only occur if $nequiv0pmod{2^j}$ for all $j$ ($implies n=0$) and equality in $(4)$ can only occur if $nequiv-1pmod{2^j}$ for all $j$ ($implies n=-1$). Therefore, for $nge1$,
$$
frac{2n}3ltsum_{k=1}^nfrac{g(k)}kltfrac{2n}3+frac23tag5
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:49










  • $begingroup$
    And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:56










  • $begingroup$
    For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
    $endgroup$
    – robjohn
    Dec 19 '18 at 21:40


















2












$begingroup$

Since $2^jmid k$ for each $jle v_2(k)$ and $sumlimits_{j=1}^n2^{-j}=1-2^{-n}$, we have
$$
2^{-v_2(k)}=1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}tag1
$$

where $[dots]$ are Iverson brackets. Thus,
$$
begin{align}
sum_{k=1}^nfrac{g(k)}k
&=sum_{k=1}^n2^{-v_2(k)}\
&=sum_{k=1}^nleft(1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}right)\
&=n-sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}tag2
end{align}
$$

Furthermore, since $leftlfloorfrac n{2^j}rightrfloorlefrac n{2^j}$ with equality iff $nequiv0pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&lesum_{j=1}^inftyfrac n{2^j}frac1{2^j}\
&=frac n3tag3
end{align}
$$

and $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$ with equality iff $nequiv-1pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&gesum_{j=1}^inftyfrac{n-(2^j-1)}{2^j}frac1{2^j}\
&=frac n3-frac23tag4
end{align}
$$

Equality in $(3)$ can only occur if $nequiv0pmod{2^j}$ for all $j$ ($implies n=0$) and equality in $(4)$ can only occur if $nequiv-1pmod{2^j}$ for all $j$ ($implies n=-1$). Therefore, for $nge1$,
$$
frac{2n}3ltsum_{k=1}^nfrac{g(k)}kltfrac{2n}3+frac23tag5
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:49










  • $begingroup$
    And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:56










  • $begingroup$
    For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
    $endgroup$
    – robjohn
    Dec 19 '18 at 21:40
















2












2








2





$begingroup$

Since $2^jmid k$ for each $jle v_2(k)$ and $sumlimits_{j=1}^n2^{-j}=1-2^{-n}$, we have
$$
2^{-v_2(k)}=1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}tag1
$$

where $[dots]$ are Iverson brackets. Thus,
$$
begin{align}
sum_{k=1}^nfrac{g(k)}k
&=sum_{k=1}^n2^{-v_2(k)}\
&=sum_{k=1}^nleft(1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}right)\
&=n-sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}tag2
end{align}
$$

Furthermore, since $leftlfloorfrac n{2^j}rightrfloorlefrac n{2^j}$ with equality iff $nequiv0pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&lesum_{j=1}^inftyfrac n{2^j}frac1{2^j}\
&=frac n3tag3
end{align}
$$

and $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$ with equality iff $nequiv-1pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&gesum_{j=1}^inftyfrac{n-(2^j-1)}{2^j}frac1{2^j}\
&=frac n3-frac23tag4
end{align}
$$

Equality in $(3)$ can only occur if $nequiv0pmod{2^j}$ for all $j$ ($implies n=0$) and equality in $(4)$ can only occur if $nequiv-1pmod{2^j}$ for all $j$ ($implies n=-1$). Therefore, for $nge1$,
$$
frac{2n}3ltsum_{k=1}^nfrac{g(k)}kltfrac{2n}3+frac23tag5
$$






share|cite|improve this answer











$endgroup$



Since $2^jmid k$ for each $jle v_2(k)$ and $sumlimits_{j=1}^n2^{-j}=1-2^{-n}$, we have
$$
2^{-v_2(k)}=1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}tag1
$$

where $[dots]$ are Iverson brackets. Thus,
$$
begin{align}
sum_{k=1}^nfrac{g(k)}k
&=sum_{k=1}^n2^{-v_2(k)}\
&=sum_{k=1}^nleft(1-sum_{j=1}^inftyleft[,2^j,middle|,k,right]2^{-j}right)\
&=n-sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}tag2
end{align}
$$

Furthermore, since $leftlfloorfrac n{2^j}rightrfloorlefrac n{2^j}$ with equality iff $nequiv0pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&lesum_{j=1}^inftyfrac n{2^j}frac1{2^j}\
&=frac n3tag3
end{align}
$$

and $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$ with equality iff $nequiv-1pmod{2^j}$,
$$
begin{align}
sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}
&gesum_{j=1}^inftyfrac{n-(2^j-1)}{2^j}frac1{2^j}\
&=frac n3-frac23tag4
end{align}
$$

Equality in $(3)$ can only occur if $nequiv0pmod{2^j}$ for all $j$ ($implies n=0$) and equality in $(4)$ can only occur if $nequiv-1pmod{2^j}$ for all $j$ ($implies n=-1$). Therefore, for $nge1$,
$$
frac{2n}3ltsum_{k=1}^nfrac{g(k)}kltfrac{2n}3+frac23tag5
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 14:00

























answered Dec 17 '18 at 6:27









robjohnrobjohn

268k27308633




268k27308633












  • $begingroup$
    I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:49










  • $begingroup$
    And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:56










  • $begingroup$
    For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
    $endgroup$
    – robjohn
    Dec 19 '18 at 21:40




















  • $begingroup$
    I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:49










  • $begingroup$
    And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
    $endgroup$
    – Vmimi
    Dec 19 '18 at 20:56










  • $begingroup$
    For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
    $endgroup$
    – robjohn
    Dec 19 '18 at 21:40


















$begingroup$
I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
$endgroup$
– Vmimi
Dec 19 '18 at 20:49




$begingroup$
I understood almost everything about your proof. In $(2)$, in the final summation I know per each $j$ you have to add bunches of $2^{-j}$. I think you had to add the amount of numbers that are divisible by $2^{-j}$, that's why it is $sum_{j=1}^inftyleftlfloorfrac n{2^j}rightrfloorfrac1{2^j}$?
$endgroup$
– Vmimi
Dec 19 '18 at 20:49












$begingroup$
And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
$endgroup$
– Vmimi
Dec 19 '18 at 20:56




$begingroup$
And in $4)$ you said that $leftlfloorfrac n{2^j}rightrfloorgefrac {n-left(2^j-1right)}{2^j}$, that's because if you let $leftlfloorfrac n{2^j}rightrfloor = x$, then $frac {n+1}{2^j} < x+1$ unless $n+1$ is divisible by $2^j$?.
$endgroup$
– Vmimi
Dec 19 '18 at 20:56












$begingroup$
For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
$endgroup$
– robjohn
Dec 19 '18 at 21:40






$begingroup$
For $(2)$: yes. For $(4)$: $leftlfloorfrac n{2^j}rightrfloorgefrac {n+1}{2^j}-1iffleftlfloorfrac n{2^j}rightrfloorgtfrac {n}{2^j}-1$, which is the same thing.
$endgroup$
– robjohn
Dec 19 '18 at 21:40













5












$begingroup$

Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $dfrac{g(k)}{k} = 2^{-v(k)}$.



So, let $S_n = displaystylesum_{k = 1}^{n}dfrac{g(k)}{k} = sum_{k = 1}^{n}2^{-v(k)}$.



Then, we have:



begin{align*}
S_{2n} &= displaystylesum_{k = 1}^{2n}2^{-v(k)}
\
&= sum_{k = 1}^{n}2^{-v(2k-1)} + sum_{k = 1}^{n}2^{-v(2k)}
\
&= sum_{k = 1}^{n}2^{-0} + sum_{k = 1}^{n}2^{-(1+v(k))}
\
&= sum_{k = 1}^{n}1 + dfrac{1}{2}sum_{k = 1}^{n}2^{-v(k)}
\
&= n + dfrac{1}{2}S_n
end{align*}



Also, $S_{2n+1} = displaystylesum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.



We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > dfrac{2}{3}$ and $S_1 < dfrac{4}{3}$.



Now, suppose that for some integer $n$, we have $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$.



Then, we have:



$$S_{2n} = dfrac{1}{2}S_n + n > dfrac{1}{2}left(dfrac{2n}{3}right) + n = dfrac{4n}{3} = dfrac{2 cdot 2n}{3}$$



$$S_{2n} = dfrac{1}{2}S_n + n < dfrac{1}{2}left(dfrac{2n}{3}+dfrac{2}{3}right) + n = dfrac{4n}{3} + dfrac{1}{3} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$$



$$S_{2n+1} = S_{2n}+1 > left(dfrac{4n}{3}right)+1 > dfrac{4n}{3}+dfrac{2}{3} = dfrac{2(2n+1)}{3}$$



$$S_{2n+1} = S_{2n}+1 < left(dfrac{4n}{3}+dfrac{1}{3}right)+1 = dfrac{2(2n+1)}{3} + dfrac{2}{3}.$$



Hence, $dfrac{2 cdot 2n}{3} < S_{2n} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$ and $dfrac{2(2n+1)}{3} < S_{2n+1} < dfrac{2(2n+1)}{3} + dfrac{2}{3}$.



So by induction, we have that $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$ for all $n$, as desired.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 2:35








  • 1




    $begingroup$
    If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:46








  • 1




    $begingroup$
    So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 9:52






  • 1




    $begingroup$
    If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 10:43








  • 1




    $begingroup$
    I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 11:10


















5












$begingroup$

Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $dfrac{g(k)}{k} = 2^{-v(k)}$.



So, let $S_n = displaystylesum_{k = 1}^{n}dfrac{g(k)}{k} = sum_{k = 1}^{n}2^{-v(k)}$.



Then, we have:



begin{align*}
S_{2n} &= displaystylesum_{k = 1}^{2n}2^{-v(k)}
\
&= sum_{k = 1}^{n}2^{-v(2k-1)} + sum_{k = 1}^{n}2^{-v(2k)}
\
&= sum_{k = 1}^{n}2^{-0} + sum_{k = 1}^{n}2^{-(1+v(k))}
\
&= sum_{k = 1}^{n}1 + dfrac{1}{2}sum_{k = 1}^{n}2^{-v(k)}
\
&= n + dfrac{1}{2}S_n
end{align*}



Also, $S_{2n+1} = displaystylesum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.



We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > dfrac{2}{3}$ and $S_1 < dfrac{4}{3}$.



Now, suppose that for some integer $n$, we have $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$.



Then, we have:



$$S_{2n} = dfrac{1}{2}S_n + n > dfrac{1}{2}left(dfrac{2n}{3}right) + n = dfrac{4n}{3} = dfrac{2 cdot 2n}{3}$$



$$S_{2n} = dfrac{1}{2}S_n + n < dfrac{1}{2}left(dfrac{2n}{3}+dfrac{2}{3}right) + n = dfrac{4n}{3} + dfrac{1}{3} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$$



$$S_{2n+1} = S_{2n}+1 > left(dfrac{4n}{3}right)+1 > dfrac{4n}{3}+dfrac{2}{3} = dfrac{2(2n+1)}{3}$$



$$S_{2n+1} = S_{2n}+1 < left(dfrac{4n}{3}+dfrac{1}{3}right)+1 = dfrac{2(2n+1)}{3} + dfrac{2}{3}.$$



Hence, $dfrac{2 cdot 2n}{3} < S_{2n} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$ and $dfrac{2(2n+1)}{3} < S_{2n+1} < dfrac{2(2n+1)}{3} + dfrac{2}{3}$.



So by induction, we have that $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$ for all $n$, as desired.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 2:35








  • 1




    $begingroup$
    If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:46








  • 1




    $begingroup$
    So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 9:52






  • 1




    $begingroup$
    If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 10:43








  • 1




    $begingroup$
    I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 11:10
















5












5








5





$begingroup$

Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $dfrac{g(k)}{k} = 2^{-v(k)}$.



So, let $S_n = displaystylesum_{k = 1}^{n}dfrac{g(k)}{k} = sum_{k = 1}^{n}2^{-v(k)}$.



Then, we have:



begin{align*}
S_{2n} &= displaystylesum_{k = 1}^{2n}2^{-v(k)}
\
&= sum_{k = 1}^{n}2^{-v(2k-1)} + sum_{k = 1}^{n}2^{-v(2k)}
\
&= sum_{k = 1}^{n}2^{-0} + sum_{k = 1}^{n}2^{-(1+v(k))}
\
&= sum_{k = 1}^{n}1 + dfrac{1}{2}sum_{k = 1}^{n}2^{-v(k)}
\
&= n + dfrac{1}{2}S_n
end{align*}



Also, $S_{2n+1} = displaystylesum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.



We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > dfrac{2}{3}$ and $S_1 < dfrac{4}{3}$.



Now, suppose that for some integer $n$, we have $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$.



Then, we have:



$$S_{2n} = dfrac{1}{2}S_n + n > dfrac{1}{2}left(dfrac{2n}{3}right) + n = dfrac{4n}{3} = dfrac{2 cdot 2n}{3}$$



$$S_{2n} = dfrac{1}{2}S_n + n < dfrac{1}{2}left(dfrac{2n}{3}+dfrac{2}{3}right) + n = dfrac{4n}{3} + dfrac{1}{3} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$$



$$S_{2n+1} = S_{2n}+1 > left(dfrac{4n}{3}right)+1 > dfrac{4n}{3}+dfrac{2}{3} = dfrac{2(2n+1)}{3}$$



$$S_{2n+1} = S_{2n}+1 < left(dfrac{4n}{3}+dfrac{1}{3}right)+1 = dfrac{2(2n+1)}{3} + dfrac{2}{3}.$$



Hence, $dfrac{2 cdot 2n}{3} < S_{2n} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$ and $dfrac{2(2n+1)}{3} < S_{2n+1} < dfrac{2(2n+1)}{3} + dfrac{2}{3}$.



So by induction, we have that $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$ for all $n$, as desired.






share|cite|improve this answer









$endgroup$



Let $v(k)$ be the largest integer $m$ such that $2^m$ divides $k$. As you noted, $dfrac{g(k)}{k} = 2^{-v(k)}$.



So, let $S_n = displaystylesum_{k = 1}^{n}dfrac{g(k)}{k} = sum_{k = 1}^{n}2^{-v(k)}$.



Then, we have:



begin{align*}
S_{2n} &= displaystylesum_{k = 1}^{2n}2^{-v(k)}
\
&= sum_{k = 1}^{n}2^{-v(2k-1)} + sum_{k = 1}^{n}2^{-v(2k)}
\
&= sum_{k = 1}^{n}2^{-0} + sum_{k = 1}^{n}2^{-(1+v(k))}
\
&= sum_{k = 1}^{n}1 + dfrac{1}{2}sum_{k = 1}^{n}2^{-v(k)}
\
&= n + dfrac{1}{2}S_n
end{align*}



Also, $S_{2n+1} = displaystylesum_{k = 1}^{2n+1}2^{-v(k)} = 2^{-v(2n+1)} + sum_{k = 1}^{2n}2^{-v(k)} = 2^{0} + S_{2n} = S_{2n} + 1$.



We can now proceed by induction. Trivially, $S_1 = 1$, so $S_1 > dfrac{2}{3}$ and $S_1 < dfrac{4}{3}$.



Now, suppose that for some integer $n$, we have $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$.



Then, we have:



$$S_{2n} = dfrac{1}{2}S_n + n > dfrac{1}{2}left(dfrac{2n}{3}right) + n = dfrac{4n}{3} = dfrac{2 cdot 2n}{3}$$



$$S_{2n} = dfrac{1}{2}S_n + n < dfrac{1}{2}left(dfrac{2n}{3}+dfrac{2}{3}right) + n = dfrac{4n}{3} + dfrac{1}{3} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$$



$$S_{2n+1} = S_{2n}+1 > left(dfrac{4n}{3}right)+1 > dfrac{4n}{3}+dfrac{2}{3} = dfrac{2(2n+1)}{3}$$



$$S_{2n+1} = S_{2n}+1 < left(dfrac{4n}{3}+dfrac{1}{3}right)+1 = dfrac{2(2n+1)}{3} + dfrac{2}{3}.$$



Hence, $dfrac{2 cdot 2n}{3} < S_{2n} < dfrac{2 cdot 2n}{3} + dfrac{2}{3}$ and $dfrac{2(2n+1)}{3} < S_{2n+1} < dfrac{2(2n+1)}{3} + dfrac{2}{3}$.



So by induction, we have that $dfrac{2n}{3} < S_n < dfrac{2n}{3} + dfrac{2}{3}$ for all $n$, as desired.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 2:12









JimmyK4542JimmyK4542

41.1k245107




41.1k245107








  • 1




    $begingroup$
    So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 2:35








  • 1




    $begingroup$
    If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:46








  • 1




    $begingroup$
    So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 9:52






  • 1




    $begingroup$
    If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 10:43








  • 1




    $begingroup$
    I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 11:10
















  • 1




    $begingroup$
    So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 2:35








  • 1




    $begingroup$
    If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 2:46








  • 1




    $begingroup$
    So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
    $endgroup$
    – marty cohen
    Dec 16 '18 at 9:52






  • 1




    $begingroup$
    If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 10:43








  • 1




    $begingroup$
    I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
    $endgroup$
    – JimmyK4542
    Dec 16 '18 at 11:10










1




1




$begingroup$
So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
$endgroup$
– marty cohen
Dec 16 '18 at 2:35






$begingroup$
So does $lim_{n to infty} S_n-dfrac{2n}{3}$ exist; if so, what is it?
$endgroup$
– marty cohen
Dec 16 '18 at 2:35






1




1




$begingroup$
If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
$endgroup$
– JimmyK4542
Dec 16 '18 at 2:46






$begingroup$
If $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right) = L$, then $displaystylelim_{n to infty}left(S_{2n} - tfrac{2 cdot 2n}{3}right) = L$ and $displaystylelim_{n to infty}left(S_{2n+1} - tfrac{2(2n+1)}{3}right) = L$ as well. But since $S_{2n+1}-tfrac{2(2n+1)}{3} = left(S_{2n}-tfrac{2 cdot 2n}{3}right) + dfrac{1}{3}$ holds for all $n$, taking the limit of both sides as $n to infty$ yields, $L = L + tfrac{1}{3}$, a contradiction. Thus, $displaystylelim_{n to infty}left(S_n - tfrac{2n}{3}right)$ doesn't exist.
$endgroup$
– JimmyK4542
Dec 16 '18 at 2:46






1




1




$begingroup$
So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
$endgroup$
– marty cohen
Dec 16 '18 at 9:52




$begingroup$
So $S_n$ seems to behave differently for even and odd $n$. How about $lim S_{2n+k}-2(2n+k)/3$ for $k=0$ a nd $1$?
$endgroup$
– marty cohen
Dec 16 '18 at 9:52




1




1




$begingroup$
If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
$endgroup$
– JimmyK4542
Dec 16 '18 at 10:43






$begingroup$
If we let $x_n = S_n - tfrac{2n}{3}$, then we have the relations $x_{2n} = tfrac{1}{2}x_n$ and $x_{2n+1} = x_{2n}+tfrac{1}{3}$. From generating the first $2^{20}$ terms of $x_n$ in MATLAB and plotting it, It appears that $displaystylelim_{n to infty}x_{2n+k}$ not exist for $k = 0,1$. Also, I conjecture that $displaystylelim_{n to infty}x_{n cdot 2^m+ k}$ does not exist for any $0 le k le 2^m-1$ and $m in mathbb{N}$.
$endgroup$
– JimmyK4542
Dec 16 '18 at 10:43






1




1




$begingroup$
I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
$endgroup$
– JimmyK4542
Dec 16 '18 at 11:10






$begingroup$
I can also conjecture that for any $m in mathbb{N}$ and $0 le k le 2^m-1$, we have $displaystyleliminflimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k)}{3 cdot 2^{m-1}}$ and $displaystylelimsuplimits_{n to infty}x_{n cdot 2^m + k} = dfrac{r_m(k) + 1}{3 cdot 2^{m-1}}$ where $r_m(k)$ is the number obtained by reversing the $m$ digit binary representation of $k$, e.g. $r_3(1) = r_3(001_2) = 100_2 = 4$.
$endgroup$
– JimmyK4542
Dec 16 '18 at 11:10




















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