Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$ 
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$

I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{n^2}{(n+k)^2}$, which I don't see how relates.
 A: Note that your infinite sum can be changed into a integration problem, following thus. 
$$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right)=
\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}\frac{1}{(1+\frac{k}{n})^2}=\int_{0}^{1} \frac{1}{(1+x)^2}\mathrm{d}x$$
One can verify that 
$$\int_{0}^{1} \frac{1}{(1+x)^2}\mathrm{d}x=1-\frac{1}{2}=\frac{1}{2}$$
A: $\frac{n}{n+2}-\frac{n}{2n+1}=n\int\limits_{n+2}^{2n+1}\frac{1}{x^2}dx<n(\frac{1}{n+1}^2+\frac{1}{n+2}^2+\dots + \frac{1}{2n}^2)< n\int\limits_{n+1}^{2n}\frac{1}{x^2}dx=\frac{n}{n+1}-\frac{n}{2n}$
Clearly the expressions on both sides go to $\frac{1}{2}$ when $n\to \infty$.
A: Just another way to do it, probably just for your curiosity.
$$\sum_{i=1}^n \frac 1{(n+i)^2}=\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)$$ where appears the first derivative of the digamma function.
Since, for large values of $x$ (see the Wikipedia page)
$$\psi^{(0)}(x) = \log(x) -\frac{1}{2 x}-\frac{1}{12
   x^2}+O\left(\frac{1}{x^3}\right)$$ 
$$\psi^{(1)}(x) =\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{6
   x^3}+O\left(\frac{1}{x^4}\right)$$ $$\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)=\frac{1}{2 n}-\frac{3}{8 n^2}+\frac{7}{48
   n^3}+O\left(\frac{1}{n^4}\right)$$ $$\sum_{i=1}^n \frac n{(n+i)^2}=n (\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1))=\frac{1}{2}-\frac{3}{8 n}+\frac{7}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
For illustartion purposes, if $n=10$, the exact value of the summation is $$\frac{502856614213805}{1083847519827072}\approx 0.4639551$$ while the above formula gives $$\frac{2227}{4800}\approx 0.4639583$$
A: I thought it might be instructive to present a solution that relies on the Euler-Maclaurin Summation Formula (EMSF).  To that end, we proceed.
Note that we can write 
$$\sum_{k=1}^{n}\frac{n}{(n+k)^2}=\sum_{k=n+1}^{2n}\frac{n}{k^2}$$
From the EMSF, we have
$$\begin{align}
\sum_{k=n+1}^{2n}\frac{n}{k^2}&=n\left(\int_n^{2n} \frac{1}{x^2}\,dx+\frac12\left(\frac{1}{(2n)^2}-\frac{1}{n^2}\right)+\frac16 \frac{\frac{-2}{(2n)^3}-\frac{-2}{n^3}}{2}-\frac{1}{30}\frac{\frac{-24}{(2n)^5}-\frac{-24}{n^5}}{24}+O\left(\frac{1}{n^7}\right)\right)\\\\
&=\frac12 -\frac{3}{8n}+\frac{7}{48n^2}-\frac{31}{960n^4}+O\left(\frac{1}{n^6}\right)
\end{align}$$

Obviously, the limit as $n\to \infty$ is $\frac12$ as expected.  And we need not have carried out terms beyond those that were of order $n^{-1}$.  However, the EMSF provides a powerful tool for analyzing a broader array of problems in which asymptotic expansions are developed. 

