$ \lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + \frac{1}{(2n)^2} \right)$ as Riemann sum? I am trying to evaluate the limit $$
\lim_{n \rightarrow \infty} n \left( \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \cdots + 
\frac{1}{(2n)^2} \right).$$
I have been trying to convert it to the Riemann sum of some integral, but have been unable to recongize what the integral should be. How should I go about solving this problem?

The polygamma functions are great to solve this kind of problems.
$$\sum_{k=1}^N \frac{1}{(x+k)^2}=\psi^{(1)}(x+1)-\psi^{(1)}(N+x+1)$$
$\psi^{(1)}(z)$ is the trigamma function, i.e.: the polygamma[1,z] function.
With $z=n$ and $N=n$ :
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\psi^{(1)}(n+1)-\psi^{(1)}(2n+1) \right)$$
The asymptotic expansion of the trigamma function is : $\psi^{(1)}(z+1)=\frac{1}{z}-\frac{2}{z^2}+O\left(\frac{1}{z^3}\right)$
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\frac{1}{n}-\frac{2}{n^2}-\frac{1}{2n}+\frac{2}{4n^2}+O\left(\frac{1}{n^3}\right)  \right) = \frac{1}{2}+\frac{3}{2n}+O\left(\frac{1}{n^2}\right)$$
$$\lim_{n \rightarrow \infty} n\sum_{k=1}^n \frac{1}{(n+k)^2}=\frac{1}{2}$$
