How to prove $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? How can I prove that $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? I noticed that in the sum, the denominator has terms that cancel out, but I'm not sure how to take advantage of that.
 A: $$\dfrac1{i(i+1)} = \dfrac{i+1-i}{i(i+1)} = \dfrac{i+1}{i(i+1)} - \dfrac{i}{i(i+1)} = \dfrac1i - \dfrac1{i+1}$$
Hence,
\begin{align}
S_n & = \sum_{i=1}^n \dfrac1{i(i+1)} = \sum_{i=1}^n \left(\dfrac1i - \dfrac1{i+1} \right)\\
& = \left(1 - \dfrac12\right) + \left(\dfrac12 - \dfrac13\right) + \left(\dfrac13 - \dfrac14\right) + \cdots +  \left(\dfrac1n - \dfrac1{n+1}\right)\\
& = 1 - \left(\dfrac12 - \dfrac12\right) - \left(\dfrac13 - \dfrac13\right) - \left(\dfrac14 - \dfrac14\right) - \cdots -  \left(\dfrac1{n-1} - \dfrac1{n-1}\right) - \dfrac1{n+1}\\
& = 1 - \dfrac1{n+1} = \dfrac{n}{n+1}
\end{align}
A: $$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}.$$
A: $$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}\Longrightarrow$$
$$\sum_{i=1}^n\frac{1}{i(i+1)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots +\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\ldots$$
A: Seems three people type faster)
I'll be a bit more specific: this is called partial fraction expansion. Rewrite $\frac{1}{k(k+1)}$ as $\frac{A}{k} + \frac{B}{k+1} = \frac{A(k+1) + Bk}{k(k+1)}$ for some constants $A$ and $B$, then equate the coefficients:
$$
A+B=0\\
A=1
$$
to get two fractions: $\frac{1}{k} - \frac{1}{k+1}$. With the summation you get a telescoping sum leaving only the first and the last terms: $1-\frac{1}{n+1}$
