<s>[INST] The convergence of the series $\frac{n}{2(n+1)(n+2)}$ using difference method Checking the convergence of the series:
$$\sum\frac{n}{2(n+1)(n+2)}$$
I proved that the series is divergent using limit comparison test with the simple harmonic series:
$$lim_{n\to\infty}\frac{n^2}{2(n+1)(n+2)}=1/2$$
The series diverges since the harmonic series diverges.
My problem is that when I try the difference method , I find that the series converges
$$\frac{n}{2(n+1)(n+2)}=\frac{1}{2}(\frac{-1}{n+1}+\frac{2}{n+2})$$
The partial sum
$$S_n=\frac{1}{2}[(\frac{-1}{2}+\frac{2}{3})+(\frac{-1}{3}+\frac{1}{2})+(\frac{-1}{4}+\frac{2}{5})+(\frac{-1}{5}+\frac{1}{3})+(\frac{-1}{6}+\frac{2}{7})+(\frac{-1}{7}+\frac{1}{4})+...+(\frac{-1}{n+1}+\frac{2}{n+2})]$$
$$S_n=0.5 [ \frac{2}{3}+\frac{2}{5}+\frac{-1}{n+1}]$$
I got the sum of the series by taking the limit of the partial sum when n goes to infinity:
$$S=lim_{n\to\infty}S_n=\frac{8}{15}$$
I would like to know what is wrong with the difference method ( telescoping series) .
 [/INST] The problem is that the terms do not telescope. While the partial fraction decomposition
$$
\frac{n}{2(n+1)(n+2)}=\frac12\left(\frac2{n+2}-\frac1{n+1}\right)
$$
is valid,
$$
\begin{align}
\sum_{n=0}^\infty\frac{n}{2(n+1)(n+2)}
&=\sum_{n=0}^\infty\frac12\left(\frac2{n+2}-\frac1{n+1}\right)\\
&=\frac12\left(\sum_{n=0}^\infty\frac1{n+2}+\color{#C00}{\sum_{n=0}^\infty\left(\frac1{n+2}-\frac1{n+1}\right)}\right)\\
&=\frac12\left(\sum_{n=0}^\infty\frac1{n+2}\color{#C00}{-1}\right)
\end{align}
$$
That is, it is divergent since it is the sum of a divergent, harmonic series and a convergent, telescoping series (in red).
</s>