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I am confused on the following series:

$$\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)} = 1$$

My calculator reveals that the answer found when evaluating this series is 1. However, I am not sure how it arrives at this conclusion. I understand that partial fractions will be used to create the following equation. I just don't understand how to proceed with the problem.

$$\sum\limits_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1$$

Sebastiano
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acampb311
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    Its a telescoping series .Write out few terms of series say from n = 1 to n=5 and see cancellations ...Hope you can do from there – godonichia Nov 18 '14 at 04:15

4 Answers4

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Write out a few terms of the series. You should see a pattern! But first consider the finite series:

$$\sum\limits_{n=1}^{m}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{m-1} - \frac{1}{m} + \frac{1}{m} - \frac{1}{m+1}.$$ This sum is telescoping, since it collapses like a telescope.

Everything is left except for the first and last term. Now what's the limit as $m\to \infty$?

Andrey Kaipov
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First consider the partial sum. Let $S_m = \displaystyle \sum_{n=1}^m \dfrac1{n(n+1)}$. We then have \begin{align} S_m & = \sum_{n=1}^m \left(\dfrac1n - \dfrac1{n+1}\right) = \sum_{n=1}^m \dfrac1n - \sum_{n=1}^m \dfrac1{n+1} = \sum_{n=1}^m \dfrac1n - \sum_{n=2}^{m+1} \dfrac1{n}\\ & = 1 + \sum_{n=2}^m \dfrac1n - \sum_{n=2}^m\dfrac1n - \dfrac1{m+1} = 1 - \dfrac1{m+1} \end{align} Now $$\sum_{n=1}^{\infty} \dfrac1{n(n+1)} = \lim_{m \to \infty} S_m = \lim_{m \to \infty} \left(1 - \dfrac1{m+1} \right) = 1$$

Adhvaitha
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This is an example of what's called a "telescoping" series: each term cancels some part of a following term, collapsing like a handheld telescope. Separate the two parts and with some clever manipulations you can get it to do so directly.

$$\begin{align}&\lim_{h\rightarrow\infty}\sum_{n=1}^{h}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\=&\lim_{h\rightarrow\infty}\left(\sum_{n=1}^{h}\frac{1}{n}-\sum_{n=1}^{h}\frac{1}{n+1}\right) \\=&\lim_{h\rightarrow\infty}\left(\sum_{n=1}^{h}\frac{1}{n}-\sum_{n=2}^{h+1}\frac{1}{n}\right) \\=&\lim_{h\rightarrow\infty}\left(\sum_{n=1}^{1}\frac{1}{n}+\sum_{n=2}^{h}\frac{1}{n}-\sum\limits_{n=2}^{h}\frac{1}{n}-\sum_{n=h+1}^{h+1}\frac{1}{n}\right) \\=&\lim_{h\rightarrow\infty}\left(\frac{1}{1}-\frac{1}{h+1}\right) \\=&1-0=1 \end{align}$$

Dan Uznanski
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Its a telescoping series .Write out few terms of series say from n = 1 to n=5 and see cancellations ...Hope you can do from there

godonichia
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