TITLE: How to calculate $\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$?
QUESTION [11 upvotes]: How can we compute the following limit:

$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$$

Mathematica gives the answer $\sqrt{e}$. However, I do not know to do it.

REPLY [2 votes]: For $1\le k\le n$, we have
$$\begin{align}
\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}
&=\int_1^{1+k/n^2}{dt\over t}-{k\over n(n+1)}\\
&=\int_0^{k/n^2}\left({1\over1+t}-{n\over n+1}\right)dt\\
&={1\over n+1}\int_0^{k/n^2}\left(1-tn\over1+t \right)dt\\
&\le{1\over n+1}\int_0^{k/n^2}dt\\
&\le{k\over n^2(n+1)}\\
&\le{1\over n(n+1)}
\end{align}$$
Note, the integral in the third line is clearly non-negative, which means the initial expression is also non-negative.  Therefore
$$0\le\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\le{1\over n(n+1)}$$
Consequently, since $1+2+\cdots+n={n(n+1)\over2}$, we have
$$0\le\sum_{k=1}^n\log\left(1+{k\over n^2}\right)-{1\over 2}=\sum_{k=1}^n\left(\log\left(1+{k\over n^2}\right)-{k\over n(n+1)}\right)\le{1\over n+1}$$
so by the Squeeze Theorem
$$\lim_{n\to\infty}\sum_{k=1}^n\log\left(1+{k\over n^2}\right)={1\over2}$$
and thus
$$\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)=e^{1/2}=\sqrt e$$