Inequality deduced from relatively prime numbers.












3












$begingroup$


If $a_n text{ and }b_n$ are relatively prime for all $n$ and



$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$



Deduce that



$$b_ngeq b_{n+1}$$



CURRENT THOUGHTS



I can show that



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



and making $b_n$ the subject



$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$



so it would suffice to show that



$$na_ngeq a_{n+1}+b_{n+1}$$



which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?






inequality prime-numbers irrational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    That series looks like it converges to an irrational.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:08






  • 1




    $begingroup$
    Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:11












  • $begingroup$
    Can't you just adapt the usual irrationality proof for $e$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:13










  • $begingroup$
    I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:17










  • $begingroup$
    @AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
    $endgroup$
    – user408906
    Dec 21 '18 at 6:19
















3












$begingroup$


If $a_n text{ and }b_n$ are relatively prime for all $n$ and



$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$



Deduce that



$$b_ngeq b_{n+1}$$



CURRENT THOUGHTS



I can show that



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



and making $b_n$ the subject



$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$



so it would suffice to show that



$$na_ngeq a_{n+1}+b_{n+1}$$



which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?






inequality prime-numbers irrational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    That series looks like it converges to an irrational.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:08






  • 1




    $begingroup$
    Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:11












  • $begingroup$
    Can't you just adapt the usual irrationality proof for $e$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:13










  • $begingroup$
    I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:17










  • $begingroup$
    @AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
    $endgroup$
    – user408906
    Dec 21 '18 at 6:19














3












3








3


1



$begingroup$


If $a_n text{ and }b_n$ are relatively prime for all $n$ and



$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$



Deduce that



$$b_ngeq b_{n+1}$$



CURRENT THOUGHTS



I can show that



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



and making $b_n$ the subject



$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$



so it would suffice to show that



$$na_ngeq a_{n+1}+b_{n+1}$$



which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?






inequality prime-numbers irrational-numbers







share|cite|improve this question











$endgroup$




If $a_n text{ and }b_n$ are relatively prime for all $n$ and



$$frac{a_n}{b_n}=frac{1}{n}+frac{1}{n(n+1)}+frac{1}{n(n+1)(n+2)}+cdots$$



Deduce that



$$b_ngeq b_{n+1}$$



CURRENT THOUGHTS



I can show that



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



and making $b_n$ the subject



$$b_n=(frac{na_n}{a_{n+1}+b_{n+1}})b_{n+1}$$



so it would suffice to show that



$$na_ngeq a_{n+1}+b_{n+1}$$



which would appear to be true in general as $n$ gets large. But other than this, I am unsure how to proceed?






inequality prime-numbers irrational-numbers




inequality prime-numbers irrational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 6:50









Alex Ravsky

42.4k32383




42.4k32383










asked Dec 21 '18 at 5:05









An Invisible CarrotAn Invisible Carrot

1018




1018












  • $begingroup$
    That series looks like it converges to an irrational.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:08






  • 1




    $begingroup$
    Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:11












  • $begingroup$
    Can't you just adapt the usual irrationality proof for $e$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:13










  • $begingroup$
    I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:17










  • $begingroup$
    @AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
    $endgroup$
    – user408906
    Dec 21 '18 at 6:19


















  • $begingroup$
    That series looks like it converges to an irrational.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:08






  • 1




    $begingroup$
    Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:11












  • $begingroup$
    Can't you just adapt the usual irrationality proof for $e$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 5:13










  • $begingroup$
    I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
    $endgroup$
    – An Invisible Carrot
    Dec 21 '18 at 5:17










  • $begingroup$
    @AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
    $endgroup$
    – user408906
    Dec 21 '18 at 6:19
















$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08




$begingroup$
That series looks like it converges to an irrational.
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:08




1




1




$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11






$begingroup$
Indeed it does, I am trying to write an elementary proof of this for my students -- if I can show this inequality then it will follow that the series is irrational. But I was hoping to not use the fact that it is irrational to prove the inequality!
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:11














$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13




$begingroup$
Can't you just adapt the usual irrationality proof for $e$?
$endgroup$
– Lord Shark the Unknown
Dec 21 '18 at 5:13












$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17




$begingroup$
I dont know what you mean by 'usual' proof, but I am currently establishing $x_n=frac{a_n}{b_n}$ where $e$ being rational would imply that all $x_n$ are rational, and then showing that $x_n$ is not rational by contradiction. The inequality $b_n geq b_{n+1}$ is all I need.
$endgroup$
– An Invisible Carrot
Dec 21 '18 at 5:17












$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19




$begingroup$
@AnInvisibleCarrot I wonder how to get $a_n$and $b_n$ when it doesn't really exists because your RHS seem to converge to an irrational?
$endgroup$
– user408906
Dec 21 '18 at 6:19










2 Answers
2






active

oldest

votes


















1












$begingroup$

I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.



i.e.



If



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.



Hope this helps!



If I am not mistaken, the next steps in this proof will be to conclude that



$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$



And we obtain a contradiction by



$$a_1>a_2>a_3>cdots >0$$



Since it is not possible to have a infinite decreasing sequence of positive integers?



Very nice!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57



















0












$begingroup$

There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
    $endgroup$
    – An Invisible Carrot
    Dec 24 '18 at 23:53










  • $begingroup$
    Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
    $endgroup$
    – Hugh Entwistle
    Dec 24 '18 at 23:57











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.



i.e.



If



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.



Hope this helps!



If I am not mistaken, the next steps in this proof will be to conclude that



$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$



And we obtain a contradiction by



$$a_1>a_2>a_3>cdots >0$$



Since it is not possible to have a infinite decreasing sequence of positive integers?



Very nice!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57
















1












$begingroup$

I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.



i.e.



If



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.



Hope this helps!



If I am not mistaken, the next steps in this proof will be to conclude that



$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$



And we obtain a contradiction by



$$a_1>a_2>a_3>cdots >0$$



Since it is not possible to have a infinite decreasing sequence of positive integers?



Very nice!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57














1












1








1





$begingroup$

I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.



i.e.



If



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.



Hope this helps!



If I am not mistaken, the next steps in this proof will be to conclude that



$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$



And we obtain a contradiction by



$$a_1>a_2>a_3>cdots >0$$



Since it is not possible to have a infinite decreasing sequence of positive integers?



Very nice!






share|cite|improve this answer









$endgroup$



I believe this statement naturally follows from the fact that $a_n$ and $b_n$ are defined to be coprime for all $n$.



i.e.



If



$$frac{a_{n+1}}{b_{n+1}}=frac{na_n-b_n}{b_n}$$



Then we must have that $b_n geq b_{n+1}$ otherwise $a_{n+1}$ and $b_{n+1}$ wouldn't be coprime.



Hope this helps!



If I am not mistaken, the next steps in this proof will be to conclude that



$$frac{a_n}{b_n}>frac{a_{n+1}}{b_{n+1}}Rightarrow a_{n+1}<a_n$$



And we obtain a contradiction by



$$a_1>a_2>a_3>cdots >0$$



Since it is not possible to have a infinite decreasing sequence of positive integers?



Very nice!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 0:21









Hugh EntwistleHugh Entwistle

841217




841217












  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57


















  • $begingroup$
    Yes this is exactly what I was looking for - it seems rather obvious now!
    $endgroup$
    – An Invisible Carrot
    Dec 26 '18 at 8:57
















$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57




$begingroup$
Yes this is exactly what I was looking for - it seems rather obvious now!
$endgroup$
– An Invisible Carrot
Dec 26 '18 at 8:57











0












$begingroup$

There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
    $endgroup$
    – An Invisible Carrot
    Dec 24 '18 at 23:53










  • $begingroup$
    Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
    $endgroup$
    – Hugh Entwistle
    Dec 24 '18 at 23:57
















0












$begingroup$

There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
    $endgroup$
    – An Invisible Carrot
    Dec 24 '18 at 23:53










  • $begingroup$
    Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
    $endgroup$
    – Hugh Entwistle
    Dec 24 '18 at 23:57














0












0








0





$begingroup$

There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.






share|cite|improve this answer









$endgroup$



There is a simple straightforward proof that for each $n$ $$S(n)=sum_{i=n}^inftyprod_{j=n}^i frac 1j$$ is irrational. Indeed, suppose to the contrary that $S(n)=frac {a_n}{b_n}$ where $a_n$ and $b_n$ are integers. Put $m=n+b_n$. Among $b_n$ consecutive numbers $n, n+1,dots, m-1$ exists one divisible by $b_n$. Then $$T(n)=n(n+1)cdotcdotscdot (m-1)S(n)$$ is an integer. On the other hand, $T(n)$ is a sum of an integer and $S(m)$. But
$$0<S(m)<sum_{i=m}^infty m^{m-i-1}=frac 1mcdotfrac{1}{1-frac 1{m}}=frac1{m-1}<1,$$
a contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '18 at 6:49









Alex RavskyAlex Ravsky

42.4k32383




42.4k32383












  • $begingroup$
    This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
    $endgroup$
    – An Invisible Carrot
    Dec 24 '18 at 23:53










  • $begingroup$
    Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
    $endgroup$
    – Hugh Entwistle
    Dec 24 '18 at 23:57


















  • $begingroup$
    This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
    $endgroup$
    – An Invisible Carrot
    Dec 24 '18 at 23:53










  • $begingroup$
    Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
    $endgroup$
    – Hugh Entwistle
    Dec 24 '18 at 23:57
















$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53




$begingroup$
This answers the question "Show that $S(n)$ is irrational" but it does not answer the question that I was asking.
$endgroup$
– An Invisible Carrot
Dec 24 '18 at 23:53












$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57




$begingroup$
Alex we are all familiar with these irrationality proofs, I think the OP wants help with the original inequality (i.e. that $b_{n+1}>b_n$)
$endgroup$
– Hugh Entwistle
Dec 24 '18 at 23:57


















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