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

If  $L<1,$ the series is absolutely convergent.

If  $L>1,$ the series is divergent.

If  $L=1,$ the test is inconclusive.

Solution:

(a)

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

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

(b)

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

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