Answer
The series converges to 1.
Work Step by Step
From the Definitions of Convergent and Divergent Series$:$
If the sequence of partial sums $\{S_{n}\}$ converges to $S$, then the series $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges. The limit $S$ is called the sum of the series.
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Using the hint,
$\displaystyle \frac{1}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}\qquad\times n(n+1)$
$1=A(n+1)+Bn$
$1=n(A+B)+A$
polynomial equality $\rightarrow$ equate the coefficients:$ \left\{\begin{array}{l}
A=1\\
A+B=
\end{array}\right.$
$A=1, B=-1$
$\displaystyle \frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}$
$\displaystyle \sum_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1})=1-\frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots$
The n-th partial sum is
$S_{n}=1-\displaystyle \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+$
$\displaystyle \qquad\qquad...+(\frac{1}{n}-\frac{1}{n+1})$
... all but the first and last terms cancel,
$S_{n}=1-\displaystyle \frac{1}{n+1}$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\lim_{n\rightarrow\infty}S_{n}=\lim_{n\rightarrow\infty}(1-\frac{1}{n+1})=1$
The series converges to 1.
