How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
 A: $$\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)\implies$$
$$\sum_{k=1}^n\frac1{k(k+3)}=\frac13\left(\frac11-\frac14+\frac12-\frac15+\frac13-\frac16+\frac14-\frac17+\ldots+\frac1n-\frac1{n+3}\right)=$$
$$\frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)\xrightarrow[n\to\infty]{}\frac13\left(1+\frac56\right)=\frac{11}{18}$$
A: \begin{align*}
  \frac{1}{n(n+3)} &= \frac{1}{3n}-\frac{1}{3(n+3)} \\
  \sum_{n=1}^{\infty} \frac{1}{n(n+3)} &=
  \sum_{n=1}^{\infty}
   \left[ \frac{1}{3n}-\frac{1}{3(n+3)} \right] \\
  &=\lim_{N\to \infty}
    \left[
      \sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)}
    \right] \\
  &=\sum_{n=1}^{3} \frac{1}{3n}-
    \lim_{N\to \infty} \sum_{n=1}^{3} \frac{1}{3(N+n)} \\
  &=\frac{11}{18}
\end{align*}
A: Using partial fraction decomposition 
$$
    \frac{1}{n(n+3)} = \frac{1}{3} \cdot \frac{1}{n} - \frac{1}{3} \cdot  \frac{1}{n+3}
$$
convince yourself that for any $f$, we have
$$
   \sum_{n=1}^{m} \left(f(n)-f(n+3)\right) = f(1) + f(2) + f(3) - f(m+3) - f(m+2) - f(m+1)
$$
and now take the limit of $m \to \infty$ getting
$$
    \sum_{n=1}^\infty \frac{1}{n(n+3)} = \frac{1}{3} \left(1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{18}
$$
A: The partial sum decomposition of the term of your serie is : $\frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}$. You recognise telescoping series : $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}=\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}=\frac{\frac{1}{3}}{1}+\frac{\frac{1}{3}}{2}+\frac{\frac{1}{3}}{3}=\frac{11}{18}$.
A: You can use double integral to calculate. Let $f(x)=\sum_{n=1}^\infty\frac{1}{n(n+3)}x^n$. Then $f(1)=\sum_{n=1}^\infty\frac{1}{n(n+3)}$. Clearly
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{n+3}x^{n-1}$$
and 
$$ (x^4f'(x))'=\sum_{n=1}^\infty x^{n+2}=\frac{x^3}{1-x}.$$
Thus
\begin{eqnarray}
f(1)&=&\int_0^1\left(\int_0^x\frac{1}{x^4}\frac{t^3}{1-t}dt\right)dx\\
&=&\int_0^1\left(\int_t^1\frac{1}{x^4}\frac{t^3}{1-t}dx\right)dt\\
&=&-\frac13\int_0^1\left(1-\frac{1}{t^3}\right)\frac{t^3}{1-t}dt\\
&=&\frac13\int_0^1(t^2+t+1)dt\\
&=&\frac{11}{18}.
\end{eqnarray}
A: A lot of these answers involve taking the limit as the term number approaches $\infty$, but this is pretty tedious sometimes, so I thought an alternative might help with the evaluation of the sum.
$$\sum_{n=1}^\infty \frac{1}{n(n+3)}=\sum_{n=1}^\infty \bigg(\frac{A}{n}+\frac{B}{n+3}\bigg)=\sum_{n=1}^\infty \bigg(\frac{1/3}{n}-\frac{1/3}{n+3}\bigg)=\frac{1}{3}\sum_{n=1}^\infty \bigg(\frac{1}{n}-\frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)$$$$=\frac13\bigg(\sum_{n=1}^3\frac{1}{n}+\sum_{n=4}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)=\frac13\sum_{n=1}^3\frac{1}{n}=\frac13\bigg(\frac11+\frac12+\frac13\bigg)=$$$$\frac13\bigg(\frac66+\frac36+\frac26\bigg)=\frac13\bigg(\frac{11}6\bigg)=\frac{11}{18} $$
