How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$?

Question

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.

Answer 1

Use

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

and you get a telescoping sum.

Answer 2

$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$, therefore $$\sum_{n=1}^{N} \frac{1}{n(n+1)}= 1- \frac{1}{N+1}$$

So, $$\lim_{N \to \infty} (1 - \frac{1}{N+1}) = 1.$$

Answer 3

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

$$ \tag 4 \sum_{n=0}^N \frac{1}{(n+1)(n+2)} = \sum_{n=1}^{N+1} \frac{1}{n(n+1)} = \frac{N+1}{N+2} \to 1, \,\, \text{ for }N \to \infty. $$

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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 $$ \sum_{n \neq 0,-1} \frac{1}{n(n+1)} = 2 \frac{1}{n(n+1)} = 2.$$

Answer 4

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}

Answer 5

$$ \frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1} $$ So $$ \begin{align}\sum_n^\infty \frac{1}{n(n+1)} &= \left( \frac{1}{1}-\frac{1}{2} \right) + \left( \frac{1}{2}-\frac{1}{3} \right) + \cdots\\ &= \frac{1}{1} + \left( -\frac{1}{2}+\frac{1}{2} \right) + \left( -\frac{1}{3}+\frac{1}{3} \right) + \cdots \\ &=1+ 0 + 0 + 0 + \cdots = 1 \end{align} $$

Answer 6

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 $m=1$.