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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}.}

Final Answer:
(a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{8}{3}}
(b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}}

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