Riemann sum of $\int_1^2 {1\over x^2} dx$.

There's no easy closed form for $$\sum_{k = 1}^{n} \frac{1}{(n+k)^2}\,,$$ but since we're interested in a limit we can achieve our goal by approximating the terms of the sum in such a way that the approximation has an easy closed form. A very good approximation is obtained by \begin{align} \sum_{k = 1}^{n} \frac{1}{(n+k)^2 - \frac{1}{4}} &= \sum_{k = 1}^{n} \frac{1}{\bigl(n+k - \frac{1}{2}\bigr)\bigl(n + k + \frac{1}{2}\bigr)} \\ &= \sum_{k = 1}^{n} \biggl(\frac{1}{n+k - \frac{1}{2}} - \frac{1}{n + k + \frac{1}{2}}\biggr) \\ &= \frac{1}{n + \frac{1}{2}} - \frac{1}{2n + \frac{1}{2}} \end{align} from which $$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{(n+k)^2 - \frac{1}{4}} = \frac{1}{2}$$ is easily read off.

It remains to verify that the error introduced by approximating the terms doesn't influence the result. One can argue that this is also a Riemann sum for the integral (choose the points $\xi_k = \frac{1}{n} \sqrt{(n+k)^2 - \frac{1}{4}}$ to evaluate the function at), but a direct estimate is more transparent: $$0 < \frac{1}{(n+k)^2 - \frac{1}{4}} - \frac{1}{(n+k)^2} = \frac{1}{(n+k)^2\bigl(4(n+k)^2-1\bigr)} < \frac{1}{4n^4}\,,$$ so the total difference is $$0 < n\sum_{k = 1}^n \biggl(\frac{1}{(n+k)^2 - \frac{1}{4}} - \frac{1}{(n+k)^2}\biggr) < n\cdot n\cdot \frac{1}{4n^4} = \frac{1}{4n^2}$$ and $$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{(n+k)^2} = \frac{1}{2}$$ is proved.