Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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



share|cite|improve this question
@OP Why are you leaving this page in the mess it is in? – Did Jan 10 '14 at 9:27
@Did let me see what I can do ... – Chris's sis the artist Jan 10 '14 at 11:20

1 Answer 1

up vote 14 down vote accepted



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

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

$$\therefore P=e$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.