Determine the value of the series below

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

$\sum^{\infty}_{n=1}(\frac{4}{n^2+4n+3}-9^{-n+2}4^{n+1})=\frac{5}{12}-\frac{1296}{5}$ 

$\frac{-3863}{15}$