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

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?

Answer

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.

Refer also to the related