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

I've got a limit which puzzle me several days. The question is

$$\lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$

Can you help me? Thank you in advance

Answer

Intuitively, we have

$$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$

and therefore the log-limit is $\frac{1}{2}$.

Here is a more elementary approach: Let $P_n$ denote the sequence inside the limit. Then just note that

$$P_n^2 = \left[ \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) \right]^2 = \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right)\left( 1 + \frac{n-k}{n^2} \right) = \prod_{k=1}^{n} \left( 1 + \frac{1}{n}+\frac{k(n-k)}{n^4} \right).$$

Now fix $m$ and let $n \geq m$. Since $k (n-k) \leq \frac{1}{4}n^2$, we have

$$\frac{k(n-k)}{n^4} \leq \frac{1}{4n^2} \leq \frac{1}{4mn}.$$

Thus we have

$$\left( 1 + \frac{1}{n} \right)^n \leq P_n^2 \leq \left( 1 + \frac{1+(1/4m)}{n} \right)^n.$$

Thus taking $n \to \infty$,

$$e \leq \liminf_{n\to\infty} P_n^2 \leq \limsup_{n\to\infty} P_n^2 \leq e^{1+1/(4m)}.$$

Since $m$ is now arbitrary, we have $P_n^2 \to e$, or equivalently, $P_n \to \sqrt{e}$.