$ \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?
 A: 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 : $\psi^{(1)}(z+1)=\frac{1}{z}-\frac{2}{z^2}+O\left(\frac{1}{z^3}\right)$
$$n\sum_{k=1}^n \frac{1}{(n+k)^2}=n\left(\frac{1}{n}-\frac{2}{n^2}-\frac{1}{2n}+\frac{2}{4n^2}+O\left(\frac{1}{n^3}\right)  \right) = \frac{1}{2}+\frac{3}{2n}+O\left(\frac{1}{n^2}\right)$$
$$\lim_{n \rightarrow \infty} n\sum_{k=1}^n \frac{1}{(n+k)^2}=\frac{1}{2}$$
A: $$\lim_{n \rightarrow \infty} n \left(\sum_{k  =1}^{n} \frac{1}{(n+k)^2}\right)
=\lim_{n \rightarrow \infty} n \left(\sum_{k  =1}^{n} \frac{1}{n^2(1+\dfrac kn)^2}\right)
\\\lim_{n \rightarrow \infty}  \left(\sum_{k  =1}^{n} \frac{n}{n^2(1+\dfrac kn)^2}\right)=\\\lim_{n \rightarrow \infty}  \left(\sum_{k  =1}^{n} \frac{1}{(1+\dfrac kn)^2}\dfrac 1n\right)=\\\int_{0}^{1}\dfrac{1}{(1+x)^2}dx=\\
\int_{0}^{1}{(1+x)^{-2}}dx=\dfrac{(1+x)^{-1}}{-1} \space [0,1]\\\dfrac{-1}{1+x}\space [0,1] =\dfrac {-1}{1+1}-( \dfrac {-1}{1+0})=\\\dfrac12$$
