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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?

user
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weilam06
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1 Answers1

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We have that

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

and

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

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

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