# Difference between revisions of "009C Sample Midterm 2, Problem 5"

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:
Ratio Test
Let  ${\displaystyle \sum a_{n}}$  be a series and  ${\displaystyle L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.}$
Then,

If  ${\displaystyle L<1,}$  the series is absolutely convergent.

If  ${\displaystyle L>1,}$  the series is divergent.

If  ${\displaystyle L=1,}$  the test is inconclusive.

Solution:

(a)

Step 1:
Assume that the power series  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  converges.
Let  ${\displaystyle R}$  be the radius of convergence of this power series.
We can use the Ratio Test to find  ${\displaystyle R.}$
Using the Ratio Test, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}.}\end{array}}}$

Since the radius of convergence of the series  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  is  ${\displaystyle R,}$  we have
${\displaystyle R={\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}}}}.}$
Step 2:
Now, we use the Ratio Test to find the radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}.}$
Using the Ratio Test, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}2^{n}x^{n+1}}{c_{n}2^{n+1}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{2c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}.}\end{array}}}$
Hence, the radius of convergence of this power series is
${\displaystyle {\frac {2}{\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}}}}=2R.}$
Therefore, this power series converges.

(b)

Step 1:
Assume that the power series  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  converges.
Let  ${\displaystyle R}$  be the radius of convergence of this power series.
We can use the Ratio Test to find  ${\displaystyle R.}$
Using the Ratio Test, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}.}\end{array}}}$

Since the radius of convergence of the series  ${\displaystyle \sum _{n=0}^{\infty }c_{n}x^{n}}$  is  ${\displaystyle R,}$  we have
${\displaystyle R={\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}}}}.}$
Step 2:
Now, we use the Ratio Test to find the radius of convergence of the series ${\displaystyle \sum _{n=0}^{\infty }c_{n}(-x)^{n}.}$
Using the Ratio Test, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}(-x)^{n+1}}{c_{n}(-x)^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}(-x)}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}.}\end{array}}}$
Hence, the radius of convergence of this power series is
${\displaystyle {\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\bigg (}{\frac {c_{n+1}}{c_{n}}}{\bigg )}}}}=R.}$
Therefore, this power series converges.