Proving divergence of the fraction of the square root












2












$begingroup$


I saw this question somewhere and was having a hard time proving it so I would love to get some help.



Prove using $varepsilon$ and $N$ that the following sequence is divergent



$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$



I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.






sequences-and-series divergent-series epsilon-delta







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  • 1




    $begingroup$
    Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
    $endgroup$
    – Wojowu
    Nov 15 '15 at 21:10










  • $begingroup$
    didn't really understand,can't really see the connection between the 2 sequences
    $endgroup$
    – user290361
    Nov 15 '15 at 21:32
















2












$begingroup$


I saw this question somewhere and was having a hard time proving it so I would love to get some help.



Prove using $varepsilon$ and $N$ that the following sequence is divergent



$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$



I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.






sequences-and-series divergent-series epsilon-delta







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
    $endgroup$
    – Wojowu
    Nov 15 '15 at 21:10










  • $begingroup$
    didn't really understand,can't really see the connection between the 2 sequences
    $endgroup$
    – user290361
    Nov 15 '15 at 21:32














2












2








2





$begingroup$


I saw this question somewhere and was having a hard time proving it so I would love to get some help.



Prove using $varepsilon$ and $N$ that the following sequence is divergent



$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$



I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.






sequences-and-series divergent-series epsilon-delta







share|cite|improve this question











$endgroup$




I saw this question somewhere and was having a hard time proving it so I would love to get some help.



Prove using $varepsilon$ and $N$ that the following sequence is divergent



$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$



I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.






sequences-and-series divergent-series epsilon-delta




sequences-and-series divergent-series epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 '16 at 18:55









gebruiker

5,00851963




5,00851963










asked Nov 15 '15 at 18:21









user290361user290361

112




112








  • 1




    $begingroup$
    Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
    $endgroup$
    – Wojowu
    Nov 15 '15 at 21:10










  • $begingroup$
    didn't really understand,can't really see the connection between the 2 sequences
    $endgroup$
    – user290361
    Nov 15 '15 at 21:32














  • 1




    $begingroup$
    Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
    $endgroup$
    – Wojowu
    Nov 15 '15 at 21:10










  • $begingroup$
    didn't really understand,can't really see the connection between the 2 sequences
    $endgroup$
    – user290361
    Nov 15 '15 at 21:32








1




1




$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10




$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10












$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32




$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.



We may always assume without loss of generality that $N>1$.



Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.



$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$

This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.

If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.



(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )



Since no $L$ will do, we must conclude that $(a_n)$ diverges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You're right. I will edit.
    $endgroup$
    – gebruiker
    Dec 1 '18 at 14:22










  • $begingroup$
    @matan129 Should work now
    $endgroup$
    – gebruiker
    Dec 1 '18 at 15:20











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1 Answer
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1 Answer
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1












$begingroup$

If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.



We may always assume without loss of generality that $N>1$.



Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.



$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$

This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.

If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.



(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )



Since no $L$ will do, we must conclude that $(a_n)$ diverges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You're right. I will edit.
    $endgroup$
    – gebruiker
    Dec 1 '18 at 14:22










  • $begingroup$
    @matan129 Should work now
    $endgroup$
    – gebruiker
    Dec 1 '18 at 15:20
















1












$begingroup$

If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.



We may always assume without loss of generality that $N>1$.



Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.



$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$

This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.

If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.



(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )



Since no $L$ will do, we must conclude that $(a_n)$ diverges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You're right. I will edit.
    $endgroup$
    – gebruiker
    Dec 1 '18 at 14:22










  • $begingroup$
    @matan129 Should work now
    $endgroup$
    – gebruiker
    Dec 1 '18 at 15:20














1












1








1





$begingroup$

If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.



We may always assume without loss of generality that $N>1$.



Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.



$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$

This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.

If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.



(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )



Since no $L$ will do, we must conclude that $(a_n)$ diverges.






share|cite|improve this answer











$endgroup$



If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.



We may always assume without loss of generality that $N>1$.



Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.



$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$

This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.

If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.



(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )



Since no $L$ will do, we must conclude that $(a_n)$ diverges.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 14:47

























answered Jan 14 '16 at 18:54









gebruikergebruiker

5,00851963




5,00851963












  • $begingroup$
    You're right. I will edit.
    $endgroup$
    – gebruiker
    Dec 1 '18 at 14:22










  • $begingroup$
    @matan129 Should work now
    $endgroup$
    – gebruiker
    Dec 1 '18 at 15:20


















  • $begingroup$
    You're right. I will edit.
    $endgroup$
    – gebruiker
    Dec 1 '18 at 14:22










  • $begingroup$
    @matan129 Should work now
    $endgroup$
    – gebruiker
    Dec 1 '18 at 15:20
















$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22




$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22












$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20




$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20


















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