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
&nbsp,   ${\frac {a_{2}}{a_{1}}}={\frac {-2}{4}},~{\frac {a_{3}}{a_{2}}}={\frac {1}{-2}},$  and  ${\frac {a_{4}}{a_{2}}}={\frac {-1}{2}}.$
So, this is a geometric series with  $r={\frac {-1}{2}}.$
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:

Step 2:

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

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