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

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 $$x-\frac{x^2}{2} \leq \ln(x+1) \leq x$$.

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

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

$$\frac{1}{2}\leq\lim_{n\to\infty}S\leq\frac{1}{2}$$

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