<s>[INST] How to analyze convergence and sum of a telescopic series? I can't find a generic form Hello I would like to know what is the generic form to calculate a telescopic series.
http://www.dummies.com/how-to/content/how-to-analyze-a-telescoping-series.html
here uses a sum that startes at $n=1$, but what if it started at $n=2$ or $n=3$?
the $n$ index has any influence over the sum? I think yes, but no one explains this..
my series is: 
$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$
how do I solve this really step by step for new people to those series
 [/INST] If a telescoping sum starts at $n=m$, then
$$
\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right) =a_{m}-a_{N+1}
$$
and the telescoping series is thus
$$
\begin{eqnarray*}
\sum_{n=m}^{\infty }\left( a_{n}-a_{n+1}\right)  &=&\lim_{N\rightarrow
\infty }\sum_{n=m}^{N}\left( a_{n}-a_{n+1}\right)  \\
&=&a_{m}-\lim_{N\rightarrow \infty }a_{N+1}=a_{m}-\lim_{N\rightarrow \infty
}a_{N+1} \\
&=&a_{m}-\lim_{N\rightarrow \infty }a_{N}.
\end{eqnarray*}
$$
Of course the series converges if and only if there exists $\lim_{N\rightarrow \infty }a_{N}.$
The case $m=1$ is
$$
\begin{eqnarray*}
\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)  &=&a_{1}-a_{N+1} \\
\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)  &=&a_{1}-\lim_{N\rightarrow
\infty }a_{N}.
\end{eqnarray*}
$$
The difficult part is to write a series $
\sum_{n=m}^{\infty}u_n$ in the form  $
\sum_{n=m}^{\infty}\left( a_{n}-a_{n+1}\right)$, when possible. Concerning Wilf-Zeilberger method (also called "creative telescoping") see answers to the question Mathematical Telescoping.
Example: write $\frac{2n+1}{n^{2}\left( n+1\right) ^{2}}=\frac{1}{n^{2}}-
\frac{1}{\left( n+1\right) ^{2}}$ to evaluate 
$$
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{2n+1}{n^{2}\left( n+1\right) ^{2}}
&=&\sum_{n=1}^{\infty }\left( \frac{1}{n^{2}}-\frac{1}{\left( n+1\right) ^{2}
}\right)  \\
&=&1-\lim_{N\rightarrow \infty }\frac{1}{N^{2}}=1-0=1.
\end{eqnarray*}
$$
Added in response to the edited question. The standard technique is to
expand $\frac{4}{(n+1)(n+2)}$ into partial fractions. Write
$$
\begin{eqnarray*}
\frac{4}{(n+1)(n+2)} &=&\frac{A}{n+1}+\frac{B}{n+2} \\
&=&\frac{(n+2)A+(n+1)B}{(n+1)(n+2)} \\
&=&\frac{\left( A+B\right) n+2A+B}{(n+1)(n+2)}
\end{eqnarray*}
$$
and find the constants $A$ and $B$. The following system must hold 
$$
\left\{ 
\begin{array}{c}
A+B=0 \\ 
2A+B=4
\end{array}
\right. \Leftrightarrow \left\{ 
\begin{array}{c}
A=4 \\ 
B=-4.
\end{array}
\right. 
$$
So
$$
\frac{4}{(n+1)(n+2)}=\frac{4}{n+1}-\frac{4}{n+2}
$$
and, since $a_{n}=\frac{4}{n+1}$, you get
$$
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{4}{(n+1)(n+2)} &=&\sum_{n=1}^{\infty }\left( \frac{
4}{n+1}-\frac{4}{n+2}\right)  \\
&=&\frac{4}{1+1}-\lim_{n\rightarrow \infty }\frac{4}{n+1} \\
&=&2-0=2.
\end{eqnarray*}
$$
</s>