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)  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}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3^{2}}-\frac{1}{2\cdot3^{3}}+\frac{1}{2\cdot3^{4}}-\frac{1}{2\cdot3^{5}}+\cdots }

(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 \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}}

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Background Information:
1. For a geometric series 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 \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,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
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 \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  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 s_1,s_2,s_3,}   we notice a pattern.
We have
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 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
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 \lim_{n\rightarrow \infty} s_n.}
We have
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 \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  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{3}{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{3}{8}}
(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{3}{2}}

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