009C Sample Midterm 2, Problem 5

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.

Final Answer:
(a)     The series converges.
(b)     The series converges.

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