Suppose $p> 1$ and the sequence $\{x_n\}_{n=1}^{\infty}$ has a general term of $$x_n=\prod\limits^{n}_{k=1}{\left(1+\frac{k}{n^p}\right)} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space n=1,2,3, \cdots$$ Show that the sequence $\{x_n\}_{n=1}^{\infty}$ converges, and hence find $$\lim_{n\rightarrow\infty}{x_n}$$ 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?

  • $\begingroup$ There's no way it converges when $p=1$ $\endgroup$ – mathworker21 Nov 22 '18 at 14:19
  • $\begingroup$ What if the case for $p>1$? I have edited the question. $\endgroup$ – weilam06 Nov 22 '18 at 14:21
  • $\begingroup$ $\log(x_n) \le \sum_{k=1}^n \log(1+\frac{k}{n^p}) \le \sum_{k=1}^n \frac{k}{n^p} = \frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p \le 2$. $\endgroup$ – mathworker21 Nov 22 '18 at 14:23

We have that

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


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

therefore the sequence converges for $p\ge 2$

  • 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

  • $\begingroup$ How do you attempt $\sum\limits^{n}_{k=1}{\log {(1+\frac{k}{n^p})}^{n^p}}$~$\frac{n^2}{2}$? $\endgroup$ – weilam06 Nov 22 '18 at 15:06
  • $\begingroup$ For n large $(1+k/n^p)^{n^p}\to e^k$ and $\sum k=n(n+1)/2$. $\endgroup$ – gimusi Nov 22 '18 at 15:23
  • $\begingroup$ @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. $\endgroup$ – gimusi Nov 22 '18 at 15:56
  • $\begingroup$ That's okay. I understand that $\frac{1}{2n^{p-2}}$ is a $p-$series. $\endgroup$ – weilam06 Nov 22 '18 at 16:08
  • $\begingroup$ @weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye $\endgroup$ – gimusi Nov 22 '18 at 16:13

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