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

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