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

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.

## marked as duplicate by dustin, Did, Lord_Farin, davidlowryduda♦Feb 24 '15 at 22:00

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}

• I like this one! Do you have some method to find such functions with the terms of the series as coefficients? – glS Feb 24 '15 at 20:35
• @glance I like it too ;-) This is how one could start: recalling $log(1+x)=x+\frac{x^2}{2}+\frac{x^3}{3}$ hence I have the division by $n$, now $\frac{log(1+x)}{x}=1+\frac{x}{2}+\frac{x^2}{3}$ hence we have division by "n+1" the rest is manipulations ... – Math-fun Feb 24 '15 at 20:56

Use

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

and you get a telescoping sum.

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

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

• should be $1 - \frac{1}{N+1}$ instead – spin Feb 24 '15 at 19:16

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

• But for $k=1$, the sum is of $\dfrac1{(n+1)(n+2)}$, not $\dfrac1{n(n+1)}$. – marty cohen Feb 24 '15 at 19:47
• @martycohen thanks for the tip. Now it should be correct – glS Feb 24 '15 at 20:28
• Shouldn't that be $\dfrac{N+2}{N+3}$ since the sum goes to $N+1$? – marty cohen Feb 25 '15 at 3:10
• @martycohen I wouldn't say so. The first term of (4) is identical to (3) with $k=1$. The second term is just the first one with a change of summation index. – glS Feb 25 '15 at 8:39

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

• Your solution bugs me because it reminds me of this: $$\sum_{n=1}^{\infty}(-1)^n = -1+1-1+1-1+\cdots = (-1+1)+(-1+1)+(-1+1)+\cdots = 0+0+0+\cdots = 0.$$ – eeeeeeeeee Feb 24 '15 at 19:06
• Indeed, here it's important that the each of the terms $\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$. Otherwise telescoping doesn't make sense – spin Feb 24 '15 at 19:15

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 $m=1$.

• @glance whoops! That $k$ should be an $m$. Sorry about posting the same thing as you; on first glance, your induction proof looked to be a slightly different result. – graydad Feb 24 '15 at 20:32
• no worries. I like your generalized formula anyway! – glS Feb 24 '15 at 20:37