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)  $\sum _{n=0}^{\infty }c_{n}(-x)^{n}$

ExpandFoundations:
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)

ExpandStep 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}$
Since the radius of convergence of the series  $\sum _{n=0}^{\infty }c_{n}x^{n}$  is  $R,$  we have
$R={\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}.$

ExpandStep 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}2^{n}x^{n+1}}{c_{n}2^{n+1}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{2c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {{\frac {|x|}{2}}\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}$
Hence, the radius of convergence of this power series is
${\frac {2}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}=2R.$
Therefore, this power series converges.

(b)

ExpandStep 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}x}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}$
Since the radius of convergence of the series  $\sum _{n=0}^{\infty }c_{n}x^{n}$  is  $R,$  we have
$R={\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}.$

ExpandStep 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}(-x)^{n+1}}{c_{n}(-x)^{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {c_{n+1}(-x)}{c_{n}}}{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}$
Hence, the radius of convergence of this power series is
${\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}=R.$
Therefore, this power series converges.

ExpandFinal Answer:
(a)     converges
(b)     converges

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