Sum of special series: $1/(1\cdot2) + 1/(2\cdot3 )+ 1/(3\cdot4)+\cdots$ 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?
 A: 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,
A: Hint: Write the nth term as $\frac{1}{n}-\frac{1}{n+1}$.
A: 
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
$$
\lim_{n \to \infty} \left(1 - \frac 1{n+1}\right) \to 1.
$$
