This question already has an answer here:

Problem: find $\lim_{n \to \infty}\left(1+\frac{1}{n^2}\right)\left( 1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right)$

My thought: Taking the log and we get $\sum_{k=1}^{n}\ln(1+\frac{k}{n^{2}})$. There is an inequality in another problem, which says: $$\frac{k}{n+k}\lt \ln\left(1+\frac{k}{n}\right)\lt\frac{k}{n}, \forall k \in N^{+} $$. So I think maybe I can use it here. Plug in $n^2$ and sum over k, we get $$\sum_{k=1}^{n}\frac{k}{n^2+k}\lt \sum_{k=1}^{n}\ln\left(1+\frac{k}{n^{2}}\right) \lt\sum_{k=1}^{n}\frac{k}{n^2}=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{n+1}{2n}$$ Anyway, the limit is $\sqrt{e}$(So if we take the limit of both sides of the above inequality, it should be$\frac{1}{2}$ and the right hand side is just right), but I cannot go further with the left side.

Any hint would be appreciated!


marked as duplicate by Alex M., Lee Mosher, Carsten S, Qwerty, Ng Chung Tak Dec 16 '16 at 21:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have you tried the Stirling formula ? Your product is the same as $\frac{(n(n+1))!}{(n^2)! n^{2n}}$. $\endgroup$ – user90369 Dec 16 '16 at 10:57
  • $\begingroup$ Is the answer $\frac{1}{2}$? $\endgroup$ – Shraddheya Shendre Dec 16 '16 at 11:02
  • $\begingroup$ It's $\sqrt{e}$. $\endgroup$ – Han Tang Dec 16 '16 at 11:07
  • $\begingroup$ @user90369, do you mean using Stirling formula to approximate both $(n(n+1))!$ and $(n^2)!$ ? $\endgroup$ – Han Tang Dec 16 '16 at 11:13
  • 6
    $\begingroup$ Duplicate: $\displaystyle\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)$. (Found using Approach0.xyz) $\endgroup$ – Workaholic Dec 16 '16 at 12:22

In the lower sum, estimate $k/(n^2+k)$ by $k/(n^2+n)$ and you are done.

  • $\begingroup$ Aha! I got it ! $\endgroup$ – Han Tang Dec 16 '16 at 11:45
  • $\begingroup$ Very nice . :-) $\endgroup$ – user90369 Dec 16 '16 at 12:25


Since $1\le k\le n,$

$$\frac{k}{n^2+k}\ge\frac{k}{n^2+n}\implies \sum_{k=1}^n\frac{k}{n^2+k}\ge\sum_{k=1}^n\frac{k}{n^2+n}=\frac1{n^2+n}\sum_{k=1}^nk$$

Alternative Approach:


Consider $$\lim_{n\to\infty}(1+\frac k{n^2})^{n^2}=e^k$$

$$\lim_{n\to\infty}\ln(1+\frac k{n^2})^{n^2}=\lim_{n\to\infty}n^2\ln(1+\frac k{n^2})=k$$

$$\lim_{n\to\infty}\ln(1+\frac k{n^2})=\lim_{n\to\infty}\frac k{n^2}$$

  • $\begingroup$ Does the alternative approach mean: I can take $n^2$ of $\left(1+\frac{1}{n^2}\right)\left( 1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right)$ to get $\left(1+\frac{1}{n^2}\right)^{n^2}\left(1+\frac{2}{n^2}\right)^{n^2}\cdots\left(1+\frac{n}{n^2}\right)^{n^2}$ and the limit is $e \cdot e^2 \cdots e^n = e^{\frac{n(n+1)}{2}}$. Then take $n^2 $ th square root to get $e^{\frac{n+1}{2n}}$, which approches $e^{\frac{1}{2}}$. $\endgroup$ – Han Tang Dec 16 '16 at 12:01
  • $\begingroup$ Yes, you are correct. $\endgroup$ – Mythomorphic Dec 16 '16 at 13:36

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