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

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.

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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 $y \rightarrow 0$.

$$\therefore P=e$$

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How does the third equality follow? – Pedro Tamaroff Feb 28 '13 at 20:08
Taylor expansion of the logs for large $n$. Note that we can do this because of the fact that the argument never exceeds $1/n$, which is small in this limit. – Ron Gordon Feb 28 '13 at 20:10
@rlgordonma: This is the way I would go. (+1) – user26872 Mar 1 '13 at 1:48