Determine the value of the following series
$\sum^{\infty}_{n=1}(\frac{4}{n^2+4n+3}-9^{-n+2}4^{n+1})$

Solution
To determine their value, we must check if the series is a convergent series. 
$\sum^{\infty}_{n=1}\frac{4}{n^2+4n+3}-\sum^{\infty}_{n=1}9^{-n+2}4^{n+1}$
We need to find the values for each part of the series. So we will begin with the telescoping series:
$\sum^{\infty}_{n=1}\frac{4}{n^2+4n+3}$
$=4\sum^{\infty}_{n=1}\frac{1}{n^2+4n+3}$
Resolve into partial fractions
$4\sum^{\infty}_{n=1}\frac{1}{n^2+4n+3}=\frac{A}{n+1}+\frac{B}{n+3}$
$=\frac{1}{2}\sum^{\infty}_{n=1}(\frac{1}{n+1}-\frac{1}{n+3})$
The partial sums for this is:
$S_n=\frac{1}{2}[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})$
$+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})$
$+(\frac{1}{6}-\frac{1}{8})+...+(\frac{1}{n-2}-\frac{1}{n})$
$+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})+(\frac{1}{n+1}-\frac{1}{n+3})$
$\lim_{n\rightarrow\infty}S_n=\frac{1}{2}[\frac{1}{2}+\frac{1}{3}+\frac{1}{n-1}-\frac{1}{n+3}]$
$=\frac{1}{2}(\frac{5}{6})=\frac{5}{12}$
Therefore, this a convergent series.

Now, we solve for the second series.
$\sum^\infty_{n=1}9^{-n+2}4^{n+1}$
$=\sum^\infty_{n=1}9^{-(n-2)}4^{n+1}$
$\sum^\infty_{n=1}\frac{4^{n+1}}{9^{n-2}}$
$=\sum^\infty_{n=1}\frac{4^{n-1}.4^2}{9^{n-1}.9^{-1}}$
$\sum^\infty_{n=1}144(\frac{4}{9})^{n-1}$
Using $S_n=\frac{a}{1-r}=\frac{144}{1-\frac{4}{9}}=\frac{1296}{5}$
This is also a convergent series. 
Hence we conclude, the sum of the series is convergent since each series is a convergent series and
$\sum^{\infty}_{n=1}(\frac{4}{n^2+4n+3}-9^{-n+2}4^{n+1})=\frac{5}{12}-\frac{1296}{5}$ 
$\frac{-3863}{15}$