Prove the convergence of $prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} $ and Find Its Limit












1















Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.




I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?






limits convergence







share|cite|improve this question
























  • There's no way it converges when $p=1$
    – mathworker21
    Nov 22 at 14:19










  • What if the case for $p>1$? I have edited the question.
    – weilam06
    Nov 22 at 14:21










  • $log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
    – mathworker21
    Nov 22 at 14:23
















1















Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.




I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?






limits convergence







share|cite|improve this question
























  • There's no way it converges when $p=1$
    – mathworker21
    Nov 22 at 14:19










  • What if the case for $p>1$? I have edited the question.
    – weilam06
    Nov 22 at 14:21










  • $log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
    – mathworker21
    Nov 22 at 14:23














1












1








1


1






Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.




I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?






limits convergence







share|cite|improve this question
















Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.




I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?






limits convergence




limits convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 14:35









gimusi

1




1










asked Nov 22 at 14:06









weilam06

1109




1109












  • There's no way it converges when $p=1$
    – mathworker21
    Nov 22 at 14:19










  • What if the case for $p>1$? I have edited the question.
    – weilam06
    Nov 22 at 14:21










  • $log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
    – mathworker21
    Nov 22 at 14:23


















  • There's no way it converges when $p=1$
    – mathworker21
    Nov 22 at 14:19










  • What if the case for $p>1$? I have edited the question.
    – weilam06
    Nov 22 at 14:21










  • $log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
    – mathworker21
    Nov 22 at 14:23
















There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19




There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19












What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21




What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21












$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23




$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23










1 Answer
1






active

oldest

votes


















3














We have that



$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$



and



$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$



therefore the sequence converges for $pge 2$




  • for $p=2 implies x_n to sqrt e$

  • for $p>2 implies x_n to 1$


and diverges for $1<p<2$.



Refer also to the related




  • Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

  • How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?






share|cite|improve this answer























  • How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
    – weilam06
    Nov 22 at 15:06












  • For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
    – gimusi
    Nov 22 at 15:23










  • @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
    – gimusi
    Nov 22 at 15:56










  • That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
    – weilam06
    Nov 22 at 16:08










  • @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
    – gimusi
    Nov 22 at 16:13











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3














We have that



$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$



and



$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$



therefore the sequence converges for $pge 2$




  • for $p=2 implies x_n to sqrt e$

  • for $p>2 implies x_n to 1$


and diverges for $1<p<2$.



Refer also to the related




  • Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

  • How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?






share|cite|improve this answer























  • How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
    – weilam06
    Nov 22 at 15:06












  • For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
    – gimusi
    Nov 22 at 15:23










  • @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
    – gimusi
    Nov 22 at 15:56










  • That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
    – weilam06
    Nov 22 at 16:08










  • @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
    – gimusi
    Nov 22 at 16:13
















3














We have that



$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$



and



$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$



therefore the sequence converges for $pge 2$




  • for $p=2 implies x_n to sqrt e$

  • for $p>2 implies x_n to 1$


and diverges for $1<p<2$.



Refer also to the related




  • Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

  • How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?






share|cite|improve this answer























  • How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
    – weilam06
    Nov 22 at 15:06












  • For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
    – gimusi
    Nov 22 at 15:23










  • @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
    – gimusi
    Nov 22 at 15:56










  • That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
    – weilam06
    Nov 22 at 16:08










  • @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
    – gimusi
    Nov 22 at 16:13














3












3








3






We have that



$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$



and



$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$



therefore the sequence converges for $pge 2$




  • for $p=2 implies x_n to sqrt e$

  • for $p>2 implies x_n to 1$


and diverges for $1<p<2$.



Refer also to the related




  • Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

  • How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?






share|cite|improve this answer














We have that



$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$



and



$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$



therefore the sequence converges for $pge 2$




  • for $p=2 implies x_n to sqrt e$

  • for $p>2 implies x_n to 1$


and diverges for $1<p<2$.



Refer also to the related




  • Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$

  • How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 at 0:17

























answered Nov 22 at 14:22









gimusi

1




1












  • How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
    – weilam06
    Nov 22 at 15:06












  • For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
    – gimusi
    Nov 22 at 15:23










  • @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
    – gimusi
    Nov 22 at 15:56










  • That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
    – weilam06
    Nov 22 at 16:08










  • @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
    – gimusi
    Nov 22 at 16:13


















  • How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
    – weilam06
    Nov 22 at 15:06












  • For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
    – gimusi
    Nov 22 at 15:23










  • @weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
    – gimusi
    Nov 22 at 15:56










  • That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
    – weilam06
    Nov 22 at 16:08










  • @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
    – gimusi
    Nov 22 at 16:13
















How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06






How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06














For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23




For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23












@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56




@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56












That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08




That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08












@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13




@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13


















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