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

Question

What would you suggest me to do for the following limit? $$\lim_{n\to\infty}\left(\frac{\prod_{k=1}^{n}\displaystyle\frac{n^2+k}{n^2-k}}{e}\right)^n$$

Thanks!

Sis.

Answer 1

First of all, I'd express the products in terms of factorials, for example $\prod_{k=1}^n(n^2+k)=\frac{(n^2+n)!}{n^2!}$. and then use Stirling's approximation: $n! ∼ \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$ ($∼$ means that the ratio tends to 1 as $n \to \infty$)

Answer 2

Write

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

$$\begin{align}\log{P} &= \lim_{n\to\infty} n \left[-1 + \sum_{k=1}^n \log{\left ( \frac{n^2+k}{n^2-k} \right ) }\right ]\\ &= \lim_{n\to\infty} n \left[-1 + \sum_{k=1}^n \left \{ \log{\left ( 1 + \frac{k}{n^2} \right )} - \log{\left ( 1 - \frac{k}{n^2} \right )} \right \}\right ]\\ &= \lim_{n\to\infty} n \left[-1 + \frac{2}{n^2}\sum_{k=1}^n k\right ]\\ &= \lim_{n\to\infty} n \left[-1 + \frac{2}{n^2} \frac{n (n+1)}{2} \right]\\ &= 1\end{align}$$

Note that, in the 3rd equality, I used the fact that $\log{(1\pm y)} \sim \pm y$ as $y \rightarrow 0$.

$$\therefore P=e$$