# 009C Sample Midterm 2, Problem 5

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If  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  converges, does it follow that the following series converges?

(a)  ${\displaystyle \sum _{n=0}^{\infty }c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}}$

(b)  ${\displaystyle \sum _{n=0}^{\infty }c_{n}(-x)^{n}}$

Foundations:
A geometric series  ${\displaystyle \sum _{n=0}^{\infty }ar^{n}}$  converges if  ${\displaystyle |r|<1.}$

Solution:

(a)

Step 1:
First, we notice that  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  is a geometric series.
We have  ${\displaystyle r=x.}$
Since this series converges,
${\displaystyle |r|=|x|<1.}$
Step 2:
The series  ${\displaystyle \sum _{n=0}c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}}$  is also a geometric series.
For this series,  ${\displaystyle r={\frac {x}{2}}.}$
Now, we notice

${\displaystyle {\begin{array}{rcl}\displaystyle {|r|}&=&\displaystyle {{\bigg |}{\frac {x}{2}}{\bigg |}}\\&&\\&=&\displaystyle {\frac {|x|}{2}}\\&&\\&<&\displaystyle {\frac {1}{2}}\end{array}}}$

since  ${\displaystyle |x|<1.}$
Since  ${\displaystyle |r|<1,}$  this series converges.

(b)

Step 1:
First, we notice that  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  is a geometric series.
We have  ${\displaystyle r=x.}$
Since this series converges,
${\displaystyle |r|=|x|<1.}$
Step 2:
The series  ${\displaystyle \sum _{n=0}^{\infty }c_{n}(-x)^{n}}$  is also a geometric series.
For this series,  ${\displaystyle r=-x.}$
Now, we notice

${\displaystyle {\begin{array}{rcl}\displaystyle {|r|}&=&\displaystyle {|-x|}\\&&\\&=&\displaystyle {|x|}\\&&\\&<&\displaystyle {1}\end{array}}}$

since  ${\displaystyle |x|<1.}$
Since  ${\displaystyle |r|<1,}$  this series converges.