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

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

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Background Information:
1. For a geometric series $\sum _{n=0}^{\infty }ar^{n}$   with   $|r|<1,$
$\sum _{n=0}^{\infty }ar^{n}={\frac {a}{1-r}}.$
2. For a telescoping series, we find the sum by first looking at the partial sum   $s_{k}$
and then calculate $\lim _{k\rightarrow \infty }s_{k}.$

Solution:

(a)

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

Step 2:
Then, the series converges to the sum
${\begin{array}{rcl}\displaystyle {S}&=&\displaystyle {\frac {a}{1-r}}\\&&\\&=&\displaystyle {\frac {\frac {1}{2}}{1-(-{\frac {1}{3}})}}\\&&\\&=&\displaystyle {\frac {\frac {1}{2}}{\frac {4}{3}}}\\&&\\&=&\displaystyle {{\frac {3}{8}}.}\end{array}}$

(b)

Step 1:
We begin by using partial fraction decomposition. Let
${\frac {3}{(2x-1)(2x+1)}}={\frac {A}{2x-1}}+{\frac {B}{2x+1}}.$
If we multiply this equation by  $(2x-1)(2x+1),$  we get
$3=A(2x+1)+B(2x-1).$
If we let  $x={\frac {1}{2}},$  we get  $A={\frac {3}{2}}.$
If we let  $x=-{\frac {1}{2}},$  we get  $B=-{\frac {3}{2}}.$
So, we have
${\begin{array}{rcl}\displaystyle {\sum _{n=1}^{\infty }{\frac {3}{(2n-1)(2n+1)}}}&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {\frac {3}{2}}{2n-1}}+{\frac {-{\frac {3}{2}}}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {1}{2n-1}}-{\frac {1}{2n+1}}.}\end{array}}$

Step 2:
Now, we look at the partial sums,  $s_{n}$  of this series.
First, we have
$s_{1}={\frac {3}{2}}{\bigg (}1-{\frac {1}{3}}{\bigg )}.$
Also, we have
${\begin{array}{rcl}\displaystyle {s_{2}}&=&\displaystyle {{\frac {3}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{2}}{\bigg (}1-{\frac {1}{5}}{\bigg )}}\end{array}}$
and
${\begin{array}{rcl}\displaystyle {s_{3}}&=&\displaystyle {{\frac {3}{2}}{\bigg (}1-{\frac {1}{3}}+{\frac {1}{3}}-{\frac {1}{5}}+{\frac {1}{5}}-{\frac {1}{7}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{2}}{\bigg (}1-{\frac {1}{7}}{\bigg )}.}\end{array}}$
If we compare  $s_{1},s_{2},s_{3},$  we notice a pattern.
We have
$s_{n}={\frac {3}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}.$

Step 3:
Now, to calculate the sum of this series we need to calculate
$\lim _{n\rightarrow \infty }s_{n}.$
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {3}{2}}{\bigg (}1-{\frac {1}{2n+1}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3}{2}}.}\end{array}}$
Since the partial sums converge, the series converges and the sum of the series is  ${\frac {3}{2}}.$

Final Answer:
(a)     ${\frac {3}{8}}$
(b)     ${\frac {3}{2}}$

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