<s>[INST] How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? How can I prove that
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$
I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
 [/INST] A simple proof by induction starts by noting that
$$ \tag 1 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{2}{k(k+2)},$$
$$ \tag 2 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} = \frac{3}{k(k+3)}.$$
By induction we then see that
$$ \tag 3 \sum_{n=0}^N \frac{1}{(k+n)(k+n+1)} = \frac{N+1}{k(k+N+1)}. $$
The special case $k=1$ gives then
$$ \tag 4 \sum_{n=0}^N \frac{1}{(n+1)(n+2)} = \sum_{n=1}^{N+1} \frac{1}{n(n+1)} = \frac{N+1}{N+2} \to 1, \,\, \text{ for }N \to \infty. $$

Here is instead a different proof using the residue theorem:
Consider the function
$$ \tag 5 f(z) \equiv \frac{\pi \cot(\pi z)}{z(z+1)}. $$
This function is meromorphic with simple poles at $z \in \mathbb{Z} \setminus \{0,-1\} $, and double poles at $z=0,1$.
Moreover, it vanishes fast enough that the contour integral of $f$ over a circle $C_R$ of radius R vanished when $R \to \infty$:
$$ \tag 6  \lim_{R \to \infty} \oint_{C_R} \frac{dz}{2\pi i} f(z) = 0.$$
On the other hand, applying the residue theorem and remembering that for each $n \in \mathbb{Z}$ the Laurent series of $\cot(\pi z)$ at first orders is
$$ \tag 7 \cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^2 z^2}{45} + \mathcal O(z^3),$$
we see that
$$ \tag 8 \lim_{R \to \infty} \oint_{C_R} \frac{dz}{2\pi i} f(z)
= \sum_{n \neq 0,-1} \frac{1}{n(n+1)} - 2. $$
Putting together (6) and (8) we thus obtain
$$ \sum_{n \neq 0,-1} \frac{1}{n(n+1)} = 2 \frac{1}{n(n+1)} = 2.$$
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