Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _{n=1}^{\infty }\:\frac{1}{n}-\frac{1}{n+3} = \frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots\right)
$$
Wich means some values are zeroed, does that mean it's a telescoping sum?
I also know that  $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} \le \sum _{n=1}^{\infty }\:\frac{1}{n^2}\:$ so the series converges by the harmonic p-series.
 A: Yes, you are correct that the sum telescopes quite nicely. The partial sum formula is$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left[\left(1+\frac 12+\cdots+\frac 1m\right)-\left(\frac 14+\frac 15+\cdots+\frac 1{m+3}\right)\right]$$Notice how anything past $\frac 14$ in the first sum is automatically canceled from the second sum. Hence, the partial sum formula is given as$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left(1+\frac 12+\frac 13-\frac 1{m+1}-\frac 1{m+2}-\frac 1{m-3}\right)$$As $m\to\infty$, the fractions containing $m$ vanish, leaving$$\sum\limits_{n\geq1}\frac 1{n(n+3)}\color{blue}{=\frac {11}{18}}$$
A: Your inequality is correct, it does converge to a value and yes, it is telescoping.
Begin by separating the fraction in the sum into two terms with partial fractions
$$\frac{1}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}$$
A: Yes, it is telescoping,
\begin{align}
\sum_{n=1}^N \frac{1}{n(n+3)}&= \frac13 \sum_{n=1}^N \left[ \frac1{n}-\frac{1}{n+3}\right]\\
&= \frac13\left[1-\frac{1}{4} +\frac12-\frac1{5}+\frac13-\frac16+\ldots +\frac1{N-2}-\frac1{N+1}+\frac1{N-1}-\frac1{N+2}+\frac1N-\frac1{N+3}\right]\\
&=\frac13\left[1+\frac12+\frac13-\frac1{N+1}-\frac{1}{N+2}-\frac1{N+3} \right]
\end{align}
Now take limit $N \to \infty$,
$$\sum_{n=1}^N \frac{1}{n(n+3)}=\frac13\left[1+\frac12+\frac13 \right]$$
A: For any positive integer $k$,
if $N > k$,
$\begin{array}\\
\sum _{n=1}^{N}\:\frac{1}{n\left(n+k\right)}
&=\sum _{n=1}^{N}\frac1{k}\left(\frac1{n}-\frac1{n+k}\right)\\
&=\frac1{k}\left(\sum_{n=1}^{N}\frac1{n}-\sum_{n=1}^{N}\frac1{n+k}\right)\\
&=\frac1{k}\left(\left(\sum _{n=1}^{k}\frac1{n}+\sum _{n=k+1}^{N}\frac1{n}\right)-\left(\sum_{n=k+1}^{k+N}\frac1{n}\right)\right)\\
&=\frac1{k}\left(\left(\sum _{n=1}^{k}\frac1{n}+\sum _{n=k+1}^{N}\frac1{n}\right)-\left(\sum_{n=k+1}^{N}\frac1{n}+\sum_{n=N+1}^{N+k}\frac1{n}\right)\right)\\
&=\frac1{k}\left(\left(\sum_{n=1}^{k}\frac1{n}\right)-\left(\sum_{n=N+1}^{N+k}\frac1{n}\right)\right)\\
&\to \frac1{k}\sum_{n=1}^{k}\frac1{n}
\qquad\text{as } N \to \infty 
\text{ since } \frac{k}{N+k}\le \sum_{n=N+1}^{N+k}\frac1{n} \le \frac{k}{N+1}\\
\end{array}
$
Your case is
$k=3$.
A: To show that the series is convergent it sufficies to observe that
$$
\lim_{n\to\infty}\frac{1/[n(n+3)]}{1/n^2}=1
$$
and since $0\le\sum_n n^{-2}<\infty$, the series in question is convergent by the limit comparison test. 
