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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x={\frac {-1}{2}},}   we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle B={\frac {-1}{2}}.}
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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,  $s_{n}$  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  ${\frac {1}{2}}.$

Final Answer:
(a)    ${\frac {8}{3}}$
(b)    ${\frac {1}{2}}$

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