Answer
Converges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1 }{n (n+1)}$$
Since $f(n)= \frac{1 }{n (n+1)},\ \ f'(n) =-\frac{2n+1}{n^2\left(n+1\right)^2}$, is positive, decreasing and continuous for $n\geq 1$, then by using the integral test
\begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1 }{n (n+1)}d n \\
&= \lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1 }{n }-\frac{1 }{n +1}d n \\
&= \left.\lim _{t \rightarrow \infty}\ln\frac{n }{n+1}\right|_{1} ^{t} \\
&= \lim _{t \rightarrow \infty}\ln\frac{t}{t+1} -\ln\frac{1}{2}\\
&= \ln2 \end{aligned}
Thus the series converges.
