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

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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did you accept the solution below because you were able to derive the solution from it? If so, I'd like to see it, as neither the poster of that solution nor myself was able to figure it out. – Ron Gordon Mar 1 at 1:53

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 $n \to \infty$)

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 I think the common notation for $f(n)/g(n)\to 1$ as $n\to\infty$ is $f\sim g$ – Stahl Feb 28 at 19:51 Can you be a little more explicit about how you'd go about this? – Ron Gordon Feb 28 at 21:09 After many minutes of calculations, I give up. (I still think it's possible). The expression takes the form $\left[ \frac{1}{e} \frac{(n^2+n)! (n^2-n-1)!}{n^2! (n^2-1)!} \right]^n$, Striling's formula gives the approximation $\left[ \frac{1}{e} \frac{\sqrt{2 \pi (n^2+n)} (\frac{n^2+n}{e})^{n^2+n} \sqrt{2 \pi (n^2-n-1)} (\frac{n^2-n-1}{e})^{n^2-n-1}}{\sqrt{2 \pi n^2} (\frac{n^2}{e})^{n^2} \sqrt{2 \pi (n^2-1)} (\frac{n^2-1}{e})^{n^2-1}} \right]^n$. I'd suggest factoring this into 3-4 factors with known limits (?) – Panda Feb 28 at 22:32

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? – Peter Tamaroff Feb 28 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 at 20:10
@rlgordonma: This is the way I would go. (+1) – oen Mar 1 at 1:48