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)     converges (by the geometric series test)
(b)     converges (by the geometric series test)

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