<s>[INST] 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.
 [/INST] 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. 

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