# 009C Sample Final 2, Problem 2

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

(a)  $4-2+1-{\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{8}}+\cdots$ (b)  $\sum _{n=1}^{\infty }{\frac {1}{(2n-1)(2n+1)}}$ Foundations:
1. The sum of a convergent geometric series is   ${\frac {a}{1-r}}$ where  $r$ is the ratio of the geometric series
and  $a$ is the first term of the series.
2. The  $n$ th partial sum,  $s_{n}$ for a series  $\sum _{n=1}^{\infty }a_{n}$ is defined as

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

(a)

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

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

Step 1:
We begin by using partial fraction decomposition. Let
${\frac {1}{(2x-1)(2x+1)}}={\frac {A}{2x-1}}+{\frac {B}{2x+1}}.$ If we multiply this equation by  $(2x-1)(2x+1),$ we get
$1=A(2x+1)+B(2x-1).$ If we let  $x={\frac {1}{2}},$ we get  $A={\frac {1}{2}}.$ If we let  $x=-{\frac {1}{2}},$ we get  $B=-{\frac {1}{2}}.$ So, we have
${\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{(2n-1)(2n+1)}}}&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {\frac {1}{2}}{2n-1}}+{\frac {-{\frac {1}{2}}}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {1}{2n-1}}-{\frac {1}{2n+1}}.}\end{array}}$ Step 2:
Now, we look at the partial sums,  $s_{n}$ of this series.
First, we have
$s_{1}={\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}{\bigg )}.$ Also, we have
${\begin{array}{rcl}\displaystyle {s_{2}}&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{5}}{\bigg )}}\end{array}}$ and
${\begin{array}{rcl}\displaystyle {s_{3}}&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{5}}-{\frac {1}{7}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}1-{\frac {1}{7}}{\bigg )}.}\end{array}}$ If we compare  $s_{1},s_{2},s_{3},$ we notice a pattern.
We have
$s_{n}={\frac {1}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}.$ Step 3:
Now, to calculate the sum of this series we need to calculate
$\lim _{n\rightarrow \infty }s_{n}.$ We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {1}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$ Since the partial sums converge, the series converges and the sum of the series is  ${\frac {1}{2}}.$ (a)    ${\frac {8}{3}}$ (b)    ${\frac {1}{2}}$ 