The question is:
Find the sum of the series $$ 1/(1\cdot 2) + 1/(2\cdot3)+ 1/(3\cdot4)+\cdots$$
I tried to solve the answer and got the $n$-th term as $1/n(n+1)$. Then I tried to calculate $\sum 1/(n^2+n)$. Can you help me?
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2The series $\sum\limits_{n=1}^{\infty}{\dfrac{1}{n(n+1)}}$ is Telescoping series – M. Strochyk Dec 10 '12 at 08:52
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You have $$ \sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left( \frac 1i - \frac 1{i+1} \right) \\ = \sum_{i=1}^n \frac 1i - \sum_{i=1}^n \frac 1{i+1} \\ = \sum_{i=1}^n \frac 1i - \sum_{i=2}^{n+1} \frac 1i \\ = 1 - \frac 1{n+1}. $$ (we say that the sum telescopes). Therefore, if you let $n \to \infty$, the series converges to $$ \lim_{n \to \infty} 1 - \frac 1{n+1} = 1. $$ Hope that helps,
Patrick Da Silva
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@chndn : $1 - \frac 1{n+1} = n/(n+1)$, so there's nothing wrong there. I just thought that the limit in this form was easier to compute. – Patrick Da Silva Dec 10 '12 at 08:51
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2@Babak : Feel free to leave it there, you don't need to remove it just because I posted more stuff. – Patrick Da Silva Dec 10 '12 at 08:52
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Patrick Da Silva. I know that there is nothing wrong in your answer but I can choose this answer as correct only after 10 minutes – chndn Dec 10 '12 at 08:53
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Find the sum of the series $ \frac{1}{1\cdot 2} + \frac{1}{2\cdot3}+ \frac{1}{3\cdot4}+\cdots +\frac{1}{n(n+1)}$
You have a series called Telescoping series as M. Strochyk pointed out in the comment. Its $i^{th}$ term is $\frac{1}{i(i+1)}$ and $n^{th}$ term is $\frac{1}{n(n+1)}$ as shown in the series. Thus,
$$ \sum_{i=1}^n \frac 1{i(i+1)} = \sum_{i=1}^n \left( \frac 1i - \frac 1{i+1} \right) \\ = \sum_{i=1}^n \frac 1i - \sum_{i=1}^n \frac 1{i+1} \\ =\frac{1}{1} + \frac{1}{2}+ \frac{1}{3}+\cdots +\frac{1}{n} -\left(\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}+\cdots +\frac{1}{(n+1)}\right)\\ = 1 + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(\frac{1}{3} - \frac{1}{3}\right) + \left(\frac{1}{4} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} -\frac{1}{n}\right) - \frac{1}{(n+1)}$$ $$\therefore \ \sum_{i=1}^n \frac 1{i(i+1)} = 1 - \frac 1{n+1}$$
Therefore, if $n \to \infty$, the series converges to $1$ as $$ \lim_{n \to \infty} \left(1 - \frac 1{n+1}\right) \to 1. $$
