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?

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,

Hint: Write the nth term as $\frac{1}{n}-\frac{1}{n+1}$.