# 009C Sample Final 2, Problem 2

For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  ${\displaystyle 4-2+1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+\cdots }$

(b)  ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(2n-1)(2n+1)}}}$

Foundations:
1. The sum of a convergent geometric series is   ${\displaystyle {\frac {a}{1-r}}}$
where  ${\displaystyle r}$  is the ratio of the geometric series
and  ${\displaystyle a}$  is the first term of the series.
2. The  ${\displaystyle n}$th partial sum,  ${\displaystyle s_{n}}$  for a series  ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  is defined as

${\displaystyle s_{n}=\sum _{i=1}^{n}a_{i}.}$

Solution:

(a)

Step 1:
Let  ${\displaystyle a_{n}}$  be the  ${\displaystyle n}$th term of this sum.
We notice that
${\displaystyle {\frac {a_{2}}{a_{1}}}={\frac {-2}{4}},~{\frac {a_{3}}{a_{2}}}={\frac {1}{-2}},}$  and  ${\displaystyle {\frac {a_{4}}{a_{2}}}={\frac {-1}{2}}.}$
So, this is a geometric series with  ${\displaystyle r=-{\frac {1}{2}}.}$
Since  ${\displaystyle |r|<1,}$  this series converges.
Step 2:
Hence, the sum of this geometric series is

${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {a_{1}}{1-r}}&=&\displaystyle {\frac {4}{1-(-{\frac {1}{2}})}}\\&&\\&=&\displaystyle {\frac {4}{\frac {3}{2}}}\\&&\\&=&\displaystyle {{\frac {8}{3}}.}\end{array}}}$

(b)

Step 1:
We begin by using partial fraction decomposition. Let
${\displaystyle {\frac {1}{(2x-1)(2x+1)}}={\frac {A}{2x-1}}+{\frac {B}{2x+1}}.}$
If we multiply this equation by  ${\displaystyle (2x-1)(2x+1),}$  we get
${\displaystyle 1=A(2x+1)+B(2x-1).}$
If we let  ${\displaystyle x={\frac {1}{2}},}$  we get  ${\displaystyle A={\frac {1}{2}}.}$
If we let  ${\displaystyle x=-{\frac {1}{2}},}$  we get  ${\displaystyle B=-{\frac {1}{2}}.}$
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{(2n-1)(2n+1)}}}&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {\frac {1}{2}}{2n-1}}+{\frac {-{\frac {1}{2}}}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {1}{2n-1}}-{\frac {1}{2n+1}}.}\end{array}}}$
Step 2:
Now, we look at the partial sums,  ${\displaystyle s_{n}}$  of this series.
First, we have
${\displaystyle s_{1}={\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}{\bigg )}.}$
Also, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {s_{2}}&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{5}}{\bigg )}}\end{array}}}$
and
${\displaystyle {\begin{array}{rcl}\displaystyle {s_{3}}&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{5}}-{\frac {1}{7}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{7}}{\bigg )}.}\end{array}}}$
If we compare  ${\displaystyle s_{1},s_{2},s_{3},}$  we notice a pattern.
We have
${\displaystyle s_{n}={\frac {1}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}.}$
Step 3:
Now, to calculate the sum of this series we need to calculate
${\displaystyle \lim _{n\rightarrow \infty }s_{n}.}$
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {1}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}}$
Since the partial sums converge, the series converges and the sum of the series is  ${\displaystyle {\frac {1}{2}}.}$

(a)    ${\displaystyle {\frac {8}{3}}}$
(b)    ${\displaystyle {\frac {1}{2}}}$