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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.

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

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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}

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  • $\begingroup$ I like this one! Do you have some method to find such functions with the terms of the series as coefficients? $\endgroup$ – glS Feb 24 '15 at 20:35
  • $\begingroup$ @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 ... $\endgroup$ – Math-fun Feb 24 '15 at 20:56
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Use

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

and you get a telescoping sum.

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

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

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  • $\begingroup$ should be $1 - \frac{1}{N+1}$ instead $\endgroup$ – spin Feb 24 '15 at 19:16
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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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  • $\begingroup$ But for $k=1$, the sum is of $\dfrac1{(n+1)(n+2)}$, not $\dfrac1{n(n+1)}$. $\endgroup$ – marty cohen Feb 24 '15 at 19:47
  • $\begingroup$ @martycohen thanks for the tip. Now it should be correct $\endgroup$ – glS Feb 24 '15 at 20:28
  • $\begingroup$ Shouldn't that be $\dfrac{N+2}{N+3}$ since the sum goes to $N+1$? $\endgroup$ – marty cohen Feb 25 '15 at 3:10
  • $\begingroup$ @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. $\endgroup$ – glS Feb 25 '15 at 8:39
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$$ \frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1} $$ So $$ \begin{align}\sum_n^\infty \frac{1}{n(n+1)} &= \left( \frac{1}{1}-\frac{1}{2} \right) + \left( \frac{1}{2}-\frac{1}{3} \right) + \cdots\\ &= \frac{1}{1} + \left( -\frac{1}{2}+\frac{1}{2} \right) + \left( -\frac{1}{3}+\frac{1}{3} \right) + \cdots \\ &=1+ 0 + 0 + 0 + \cdots = 1 \end{align} $$

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    $\begingroup$ Your solution bugs me because it reminds me of this: $$ \sum_{n=1}^{\infty}(-1)^n = -1+1-1+1-1+\cdots = (-1+1)+(-1+1)+(-1+1)+\cdots = 0+0+0+\cdots = 0. $$ $\endgroup$ – eeeeeeeeee Feb 24 '15 at 19:06
  • $\begingroup$ 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 $\endgroup$ – spin Feb 24 '15 at 19:15
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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 $m=1$.

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  • $\begingroup$ @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. $\endgroup$ – graydad Feb 24 '15 at 20:32
  • $\begingroup$ no worries. I like your generalized formula anyway! $\endgroup$ – glS Feb 24 '15 at 20:37

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