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.

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 $$\frac{2227}{4800}\approx 0.4639583$$
