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:
We need to find a pattern for the partial sums in order to find a formula.
We start by calculating  $s_{2}.$  We have
$s_{2}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}.$

Step 2:
Next, we calculate  $s_{3}$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s_{4}.}   We have
${\begin{array}{rcl}\displaystyle {s_{3}}&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{3}}}-{\frac {1}{2^{4}}}{\bigg )}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{4}}}{\bigg )}}\end{array}}$
and
${\begin{array}{rcl}\displaystyle {s_{4}}&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{3}}}-{\frac {1}{2^{4}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{4}}}-{\frac {1}{2^{5}}}{\bigg )}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{5}}}{\bigg )}.}\end{array}}$

Step 3:
If we look at  $s_{2},s_{3},$  and  $s_{4},$  we notice a pattern.
From this pattern, we get the formula
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}.}

(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)     $s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}$
(b)     ${\frac {1}{2}}$

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