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.
 A: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, therefore
$$\sum_{n=1}^{N} \frac{1}{n(n+1)}= 1- \frac{1}{N+1}$$
So, $$\lim_{N \to \infty} (1 - \frac{1}{N+1}) = 1.$$
A: 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
$$ \sum_{n \neq 0,-1} \frac{1}{n(n+1)} = 2 \frac{1}{n(n+1)} = 2.$$
A: Defnie $f(x)$ as below, expand and take the limit:
\begin{align}
f(x)&=\frac{x+\log(1-x)-x \log(1-x)}{x}\\
&=\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}
\end{align}
therefore
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n(n+1)}&=\lim_{x \rightarrow 1}f(x)\\
&=\lim_{x \rightarrow 1}\frac{x+\log(1-x)-x \log(1-x)}{x}\\
&=1
\end{align}
A: Use
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
and you get a telescoping sum.
A: $$
\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}
$$
So
$$
\begin{align}\sum_n^\infty \frac{1}{n(n+1)} &= \left( \frac{1}{1}-\frac{1}{2} \right) 
+ \left( \frac{1}{2}-\frac{1}{3} \right) + \cdots\\ &= \frac{1}{1} + \left( -\frac{1}{2}+\frac{1}{2} \right) + \left( -\frac{1}{3}+\frac{1}{3} \right) + \cdots \\
&=1+ 0 + 0 + 0 + \cdots = 1
\end{align}
$$
A: I like the telescoping sum argument the best. Alternatively, you can show by induction that $$\sum_{n=1}^k\frac{1}{k(k+1)} = \frac{k}{k+1}$$ and hence in the limit, $$\lim_{k \to \infty} \left(\sum_{n=1}^k\frac{1}{k(k+1)} \right) = \lim_{k \to \infty}\left( \frac{k}{k+1}\right) = 1$$
More generally, for any integer $m \in \Bbb{N}$ then $$\sum_{n=1}^\infty\frac{1}{n(n+m)} = \frac{1}{m}\sum_{n=1}^m\frac{1}{n}$$ where you have the case of $m=1$.
