<s>[INST] 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...
 [/INST] 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}
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