Riemann sum of $\int_1^2 {1\over x^2} dx$. I've spent quite a time solving the following problem:

Evaluate using Riemann's sum:
  $$
I = \int_1^2{1\over x^2} dx
$$

I was first trying the following approach, which didn't work since the summation seems undoable to me:
$$
\Delta x = {1\over n}\\
I = \lim_{n\to\infty}\sum_{k=1}^nf\left(1+{k\over n}\right)\Delta x \\
= \lim_{n\to\infty}\sum_{k=1}^n{n^2\over (k+n)^2} {1\over n} \\
= \lim_{n\to\infty}\sum_{k=1}^n{n\over (k+n)^2}
$$
Wolfram evaluates this sum in terms of digamma function which is too advanced.
Several hours has passed before I decided to reconsider the point to choose in each partition. Let:
$$
\Delta x = {1\over n}\\
x_k = 1 + {k\over n}\\
\begin{align}
I &= \lim_{n\to\infty}\sum_{k=1}^nf\left(\sqrt{x_k x_{k-1}}\right)\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over x_k x_{k-1}}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over \left(1+{k\over n}\right)\left(1+{k-1\over n}\right)}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{n^2 \over (n+k)(n+k-1)}{1\over n}\\
&=\lim_{n\to\infty}\sum_{k=1}^n{n \over (n+k)(n+k-1)} \\
&=\lim_{n\to\infty}\sum_{k=1}^n\left({n \over (n+k-1)} - {n \over (n+k)}\right)\\
&= {n\over n} - {n\over 2n}\\
&= \boxed{{1\over 2}}
\end{align}
$$
This sum telescopes nicely. Now I'm wondering whether the first approach is even doable. I've met some other questions but the first one lists a hint I don't really understand and the second one is closed as a duplicate.
What would be the way to finish the initial approach? In the first approach, the problem is actually reduced to finding the limit which I couldn't handle. Also is there some intuition in choosing the "right" points in the partitions?

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.
