How to analyze (sum and convergence) step-by-step this series: $\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$? I have this series:
$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$
and $$\sum_{n=1}^\infty\frac{4}{(n+1)(n+3)}$$
and I would like to know how to analyze, very step-by-step.
thanks

For the second sum, take the terms $4\over (n+1)(n+3)$ and write them in the form
$$\tag{1}
{4\over (n+1)(n+3)}=  {A\over n+1}+{B\over n+3}.
$$
We need to figure out what $A$  and $B$ are. Towards that end, multiply both sides of the above by $(n+1)(n+3)$. This gives
$$\tag{2}
4=A(n+3)+B(n+1).
$$
We're presuming this holds for any value of $n$. If we set $n=-3$, then equation $(2)$ becomes
$$
4=B(-2);
$$
which tells us $B=-2$.
Similarly, if we set $n=-1$ in equation $(2)$, we obtain $A=2$. 
Thus, substituting these found  values of $A$ and $B$ into equation $(1)$, we have
$$\tag{3}
{4\over (n+1)(n+3)}=  {2\over n+1}-{2\over n+3}.
$$
The second sum in your question can then be written as
$$
\sum_{n=1}^\infty {4\over (n+1)(n+3)}=
\sum_{n=1}^\infty  \biggl( {2\over n+1}-{2\over n+3}\biggr).
$$
Now, the infinite sum $\sum\limits_{n=1}^\infty a_n $ converges if and only if the sequence of partials sums $\{S_m\}$ defined by $S_m=\sum\limits_{n=1}^m a_n $ converges. In this case, we have
$$
\sum\limits_{n=1}^\infty a_n  = \lim_{m\rightarrow\infty}\sum\limits_{n=1}^m a_n.
$$
Let's examine the $m^{\rm th}$  partial sum of our series. We explicitly write it out:
$$
\eqalign{
 \textstyle \sum\limits_{n=1}^m \Bigl[{2 \over  n+1}\!\!-\!\!  {2 \over n+3}\Bigr] 
&\!\!  = \!\!   \textstyle  \Bigl[ ( \color{orange}{2\over 2} \!\! -\!\! \color{maroon}{2\over4}  ) 
\!\! +\!\!  (\color{orange} {2\over3}\!\! -\!\!  \color{darkgreen}{2\over5} )
\!\! +\!\! (\color{maroon}{2\over4}\!\! - \!\! {2\over6})\!\! +\!\! 
(\color{darkgreen}{2\over5}\!\! - \!\! {2\over7})\!\! +\!\! 
\cdots \!\! + \!\! ({2\over (m-1)+1}\!\! -\!\! \color{pink}{2\over (m-1)+3} ) 
  \!\! +\!\!  ({2\over m+1}\!\! -\!\! \color{pink}{2\over m+3} )    \Bigr]\cr
 
}
$$
Note that the right most term in one of the parenthetical expressions cancels with the leftmost term of the second following parenthetical expression.
After canceling everything, the only terms that survive are the 
 $\color{orange}{\text{ leftmost terms of the first two parenthetical expressions}}$ and the 
 $\color{pink}{\text{ rightmost terms of the last two parenthetical expressions}}$.
So
$$
\eqalign{
  \sum\limits_{n=1}^m \Bigl[{2 \over  n+1}-  {2 \over n+3}\Bigr] 
&=  {1}+{2\over3}- {2\over (m-1)+3} -{2\over m+3}
 
}
$$
Now, taking the limit as $m\rightarrow\infty$
$$\lim_{m\rightarrow\infty}
  \sum\limits_{n=1}^m \Bigl[{2 \over  n+1}-  {2 \over n+3}\Bigr] 
=\lim_{m\rightarrow\infty}\Bigl[  1+{2\over3}- {2\over (m-1)+3} -{2\over m+3}\Bigr]
=1+{2\over3}.
$$
It follows that $\sum\limits_{n=1}^\infty {4\over (n+1)(n+3)}=1+{2\over3}$.
