To find the limit of $(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$.

First notice that $(1+\frac{1}{n^2})$ is the smallest term in the product and $(1+\frac{n}{n^2})$ is the greatest. So: $ (1+\frac{1}{n^2})^n \leq (1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2}) \leq (1+\frac{n}{n^2})^n$.

First this seemed like a good idea but then I found $ 1 \leq (1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2}) \leq e$. Which is not really helpful.

How can this problem be done?

P.S: In previous questions of the problem I had to prove that $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $x-\frac{x^2}{2} \leq \ln(x+1) \leq x$.

  • 5
    $\begingroup$ Hint: $$ \prod\limits_{k = 1}^n {\left( {1 + \frac{k}{{n^2 }}} \right)} = \exp \left( {\sum\limits_{k = 1}^n {\ln \left( {1 + \frac{k}{{n^2 }}} \right)} } \right). $$ Now use the previous questions to estimate the sum of $\ln$s. $\endgroup$
    – Gary
    Feb 6, 2022 at 12:47
  • 1
    $\begingroup$ If you have proven a result with ln, it means you should make it appear somehow $\endgroup$
    – Atmos
    Feb 6, 2022 at 12:48
  • 1
    $\begingroup$ The command of Mathematica 13 Limit[Product[1 + k/n^2, {k, 1, n}], n -> Infinity] results in $\sqrt{e}$. $\endgroup$
    – user64494
    Feb 6, 2022 at 15:20
  • $\begingroup$ Hey @Gary, thanks for the hint! But can you provide a detailed solution please? I would really appreciate that! $\endgroup$ Feb 6, 2022 at 17:55
  • $\begingroup$ @mmmmhmmmmmh There is an answer below, but you should have been able to figure out yourself. $\endgroup$
    – Gary
    Feb 6, 2022 at 22:55

1 Answer 1


You have proved that $x-\frac{x^2}{2}\leq\ln(x+1)\leq x$

Now as @Gary said $$\prod_{i=1}^n(1+\frac{i}{n^2})=e^{\sum_{i=1}^n\ln(1+\frac{k}{n^2})}$$

From $(1)$ we get , $$\sum_{i=1}^n(\frac{i}{n^2}-\frac{i^2}{2n^4})\leq\sum_{i=1}^n\ln(1+\frac{i}{n^2})\leq\sum_{i=1}^n(\frac{i}{n^2})$$

$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{12n^4}\leq S\leq\frac{n(n+1)}{2n^2}$$

where $\displaystyle S=\sum_{i=1}^n\ln(1+\frac{i}{n^2})$

Taking $\lim_{n\to\infty}$ on each side and using squeeze theorem, we get


Therefore $$\lim_{n\to\infty}\prod_{i=1}^n(1+\frac{i}{n^2})=e^{\lim_{n\to\infty}S}=\sqrt e$$


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