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

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:
From Part (a), we have
$s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}.$

Step 2:
We now calculate  $\lim _{n\rightarrow \infty }s_{n}.$
We get
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {2}{2^{2}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$

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

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