<s>[INST] show that the series converges. Using the definition, show that the series $\sum_{n=1}^\infty \frac{1}{n(n+3)}$ converges. Determine the limit.
This is what I've managed to do so far:
 $\frac{1}{n(n+3)}$ = $\frac{1}{3n}$ -$\frac{1}{3(n+3)}$
$\sum_{n=1}^\infty \frac{1}{n(n+3)}$ = $\sum_{n=1}^\infty \frac{1}{3n}$ -$\frac{1}{3(n+3)}$
S$_{k}$= $\sum_{n=1}^{k} \frac{1}{n(n+3)}$ = $\sum_{n=1}^{k} \frac{1}{3n}$ -$\frac{1}{3(n+3)}$ = $\frac{1}{3}$-$\frac{1}{12}$+$\frac{1}{6}$-$\frac{1}{15}$+$\frac{1}{9}$-$\frac{1}{18}$+$\frac{1}{12}$-$\frac{1}{21}$ $\cdots$ +$\frac{1}{3k-9}$-$\frac{1}{3k}$+$\frac{1}{3k-6}$-$\frac{1}{3k+3}$+$\frac{1}{3k-3}$-$\frac{1}{3k+6}$+$\frac{1}{3k}$-$\frac{1}{3k+9}$=$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$-$\frac{1}{3k+3
}$-$\frac{1}{3k+6}$-$\frac{1}{3k+9}$
Hence 
$\sum_{n=1}^\infty \frac{1}{n(n+3)}$ = $\lim_{k\to \infty}$S$_{k}$=$\lim_{k\to \infty}$($\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$-$\frac{1}{3k+3
}$-$\frac{1}{3k+6}$-$\frac{1}{3k+9}$)=$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$= $\frac{11}{18}$
I did the long addition just to make things clearer for myself. I was just wondering if there was anything I have missed out?
 [/INST] prove that for the partial sum is hold
$$-1/3\, \left( m+1 \right) ^{-1}-1/3\, \left( m+2 \right) ^{-1}-1/3\,
 \left( m+3 \right) ^{-1}+{\frac {11}{18}}
$$
and take then the Limit for $m$ tends to infinity
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