# 009C Sample Midterm 3, Problem 2 Detailed Solution

For each the following series find the sum, if it converges.

If you think it diverges, explain why.

(a)  ${\displaystyle {\frac {1}{2}}-{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3^{2}}}-{\frac {1}{2\cdot 3^{3}}}+{\frac {1}{2\cdot 3^{4}}}-{\frac {1}{2\cdot 3^{5}}}+\cdots }$

(b)  ${\displaystyle \sum _{n=1}^{\infty }\,{\frac {3}{(2n-1)(2n+1)}}}$

Background Information:
1. For a geometric series ${\displaystyle \sum _{n=0}^{\infty }ar^{n}}$   with   ${\displaystyle |r|<1,}$

${\displaystyle \sum _{n=0}^{\infty }ar^{n}={\frac {a}{1-r}}.}$

2. For a telescoping series, we find the sum by first looking at the partial sum   ${\displaystyle s_{k}}$

and then calculate ${\displaystyle \lim _{k\rightarrow \infty }s_{k}.}$

Solution:

(a)

Step 1:
Each term grows by a ratio of  ${\displaystyle {\frac {1}{3}}}$  and it reverses sign.
Thus, there is a common ratio  ${\displaystyle r=-{\frac {1}{3}}.}$
Also, the first term is  ${\displaystyle {\frac {1}{2}}.}$  So, we can write the series as a geometric series given by
${\displaystyle \sum _{n=0}^{\infty }\,{\frac {1}{2}}\left(-{\frac {1}{3}}\right)^{n}.}$
Step 2:
Then, the series converges to the sum

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

(b)

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

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