How to analyze convergence and sum of a telescopic series? I can't find a generic form

Solution 1:

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 $$ \begin{eqnarray*} \sum_{n=1}^{\infty }\frac{4}{(n+1)(n+2)} &=&\sum_{n=1}^{\infty }\left( \frac{ 4}{n+1}-\frac{4}{n+2}\right) \\ &=&\frac{4}{1+1}-\lim_{n\rightarrow \infty }\frac{4}{n+1} \\ &=&2-0=2. \end{eqnarray*} $$

Solution 2:

To add to Américo Tavares's answer, for more general telescoping series:

$$ \sum_{n=m}^{N}\left( a_{n}-a_{n+k}\right) =(a_{m}+a_{m+1}+\cdots+a_{m+k-1})-(a_{N+1}+a_{N+2}+\cdots+a_{N+k}) $$

$$ \begin{eqnarray*} \sum_{n=m}^{\infty }\left( a_{n}-a_{n+k}\right) &=&\lim_{N\rightarrow \infty }\sum_{n=m}^{N}\left( a_{n}-a_{n+k}\right) \\ &=&\lim_{N\rightarrow \infty }((a_{m}+a_{m+1}+\cdots+a_{m+k-1})-(a_{N+1}+a_{N+2}+\cdots+a_{N+k}))\\ &=&(a_{m}+a_{m+1}+\cdots+a_{m+k-1})-\lim_{N\rightarrow \infty }(a_{N+1}+a_{N+2}+\cdots+a_{N+k})\\ &=& a_{m}+a_{m+1}+\cdots+a_{m+k-1}-k\lim_{N\rightarrow\infty}a_N. \end{eqnarray*} $$