What would be the sum of following ?
$$\lim_{n\to\infty} \left[\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + \cdots + \frac{1}{(n+n)^{2}}\right]$$
I tried to turn it into integral :
$\displaystyle\int \frac{1}{(1+\frac{r}{n})^{2}}\frac{1}{n^{2}} $ but I can't figure out how to deal with $\frac{1}{n^{2}}$
Hint: $\frac{1}{(n+k)^2}\leq \frac{1}{n^2}$ so what is an upper bound for:
$$\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots\frac{1}{(n+n)^2}?$$
By the way, you could see this as an integral limit if you wrote it as:
$$\frac{1}{n}\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\left(1+\frac k n\right)^2}$$
But: $$\sum_{k=1}^{n} \frac{1}{n}\frac{1}{\left(1+\frac k n\right)^2}\to\int_{0}^1 \frac{dx}{(1+x)^2}=\frac12$$
So with the extra $\frac{1}{n}$, you can determine what the limit will be...
$$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{n}{(n+1)^2}$$ $$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{\frac1n}{(1+\frac1n)^2}$$ $$0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le0$$ $$\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]=0$$
The series $\sum_{j=}^{\infty}1/j^2$ converges by the Cauchy condensation test so $0<\sum_{j=n}^{j=2 n}1/j^2 < \sum_{j=n}^{\infty}1/j^2$ which $\to 0$ as $n\to \infty.$
