Limit $\lim_{n\to \infty} n(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2})$ I need some help finding the limit of the following sequence:
$$\lim_{n\to \infty}  a_n=n\left(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2}\right)$$
I can tell it is bounded by $\frac{1}{4}$ from below and decreasing from some point which tells me it is convergent.
I can't get anything else though.
So far we did not have integrals and we just started series, so this should be solved without using either.
Some hints would be very welcome :)
Thanks
 A: We can write $a_n$ in the form of the Riemann sum
$$a_n=\frac{1}{n}\sum_{k=1}^n\frac{1}{\left(1+\frac{k}{n}\right)^2}\to\int_0^1\frac{dx}{(1+x)^2}=-\frac{1}{1+x}\big|_{0}^1=\frac{1}{2}$$
A: Notice
$$\frac{1}{(n+k-1)(n+k)} \ge \frac{1}{(n+k)^2}\ge \frac{1}{(n+k)(n+k+1)}$$
We have 
$$\begin{align}a_n = n \sum_{k=1}^{n}\frac{1}{(n+k)^2}\le & n\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}=  n \sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)\\ = & n \left(\frac{1}{n}- \frac{1}{2n}\right) = \frac12\\ \text{AND}\quad \ge & n\sum_{k=1}^n\frac{1}{(n+k)(n+k+1)} =  n \sum_{k=1}^n\left(\frac{1}{n+k}-\frac{1}{n+k+1}\right)\\ = & n \left(\frac{1}{n+1} - \frac{1}{2n+1}\right) = \frac{n^2}{(n+1)(2n+1)}\end{align}$$
Since $\;\;\displaystyle \lim_{n\to\infty} \frac{n^2}{(n+1)(2n+1)} = \frac12,\;\;$ we get
$\;\;\displaystyle \lim_{n\to\infty} a_n = \frac12$.
A: $$ \lim_{n\to\infty} n\sum_{k=n+1}^{2n} \frac{1}{k^2}=\lim_{n\to\infty} \frac{1}{n}\sum_{k=n+1}^{2n}\frac{1}{\left(\frac{k}{n}\right)^2}=\int_1^2 \frac{1}{x^2} \mathrm dx=\left[-\frac{1}{x}\right]_1^2=\frac{1}{2} $$
