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?
 A: 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.
A: I've just tried one more technique while solving a similar problem and it seems to work fine. Let's split the interval $[1, 2]$ with points $x_0, x_1, \dots, x_n$ so that they form a geometric progression. Let $q$ denote the denominator of geometric progression. So the interval is split by the points: $q, q^2, \dots, q^n$, therefore:
$$
\Delta x_1 = q - 1\\
\Delta x_2 = q^2 - q\\
\cdots\\
\Delta x_n = q^n - q^{n-1}\\
$$
Now pick the points $\zeta_k$ from the rightmost point of each subsegment. Calculate the value of the function in each point $\zeta_k$:
$$
f(\zeta_k) = \left\{{1\over q^2}, {1\over q^4}, \dots, {1\over q^{2k}}\right\}
$$
Now write the sum:
$$
\begin{align}
S_n &= \sum_{k=1}^n f(\zeta_k)\Delta x_k \\
&= \sum_{k=1}^n {1\over q^{2k}} (q^k - q^{k-1}) \\
&= {1\over q^2}(q-1) + {1\over q^4}(q^2-q) + \cdots + {1\over q^{2n}}(q^n-q^{n-1}) \\
&= {1\over q^2}(q-1) + {1\over q^3}(q-1) + \cdots + {1\over q^{2n-1}}(q-1) \\
&= (q-1)\left({1\over q^2} + {1\over q^3} + {1\over q^4} + \cdots + {1\over q^{n+1}}\right)
\end{align}
$$
By geometric sum:
$$
S_n = {q^n - 1 \over q^{n + 1}}
$$
Remember $q = \sqrt[n]{2}$. Now the only part left is taking the limit:
$$
\begin{align}
I &= \lim_{n\to\infty} S_n \\
&= \lim_{n\to\infty} {q^n - 1 \over q^{n + 1}}\\
&= \lim_{n\to\infty} {(\sqrt[n]{2})^n - 1 \over (\sqrt[n]{2})^{n + 1}}\\
&= \boxed{{1\over 2}}
\end{align}
$$
It looks like this can be generalized for:
$$
\int_a^b {1\over x^k}\mathop{dx}
$$
