# 009C Sample Midterm 2, Problem 5

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

(a)  $\sum _{n=0}^{\infty }c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}$ (b)  $\sum _{n=0}^{\infty }c_{n}(-x)^{n}$ Foundations:
A geometric series  $\sum _{n=0}^{\infty }ar^{n}$ converges if  $|r|<1.$ Solution:

(a)

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

${\begin{array}{rcl}\displaystyle {|r|}&=&\displaystyle {{\bigg |}{\frac {x}{2}}{\bigg |}}\\&&\\&=&\displaystyle {\frac {|x|}{2}}\\&&\\&<&\displaystyle {\frac {1}{2}}\end{array}}$ since  $|x|<1.$ Since  $|r|<1,$ this series converges.

(b)

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

${\begin{array}{rcl}\displaystyle {|r|}&=&\displaystyle {|-x|}\\&&\\&=&\displaystyle {|x|}\\&&\\&<&\displaystyle {1}\end{array}}$ since  $|x|<1.$ Since  $|r|<1,$ this series converges.