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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!

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

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

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:

Hint:

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

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

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

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  • $\begingroup$ 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

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