# 009C Sample Final 2, Problem 4

(a) Find the radius of convergence for the power series

${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\frac {x^{n}}{n}}.}$

(b) Find the interval of convergence of the above series.

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:
We use the Ratio Test to determine the radius of convergence.
We have

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

Step 2:
The Ratio Test tells us this series is absolutely convergent if  ${\displaystyle |x|<1.}$
Hence, the Radius of Convergence of this series is  ${\displaystyle R=1.}$

(b)

Step 1:
First, note that  ${\displaystyle |x|<1}$  corresponds to the interval  ${\displaystyle (-1,1).}$
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  ${\displaystyle R=1.}$
Step 2:
First, let  ${\displaystyle x=1.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n}}.}$
This is an alternating series.
Let  ${\displaystyle b_{n}={\frac {1}{n}}.}$.
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{n+1}}<{\frac {1}{n}}}$
for all  ${\displaystyle n\geq 1.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n}}=0.}$
Therefore, this series converges by the Alternating Series Test
and we include  ${\displaystyle x=1}$  in our interval.
Step 3:
Now, let  ${\displaystyle x=-1.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n}}.}$
This is a  ${\displaystyle p}$-series with  ${\displaystyle p=1.}$  Hence, the series diverges.
Therefore, we do not include  ${\displaystyle x=-1}$  in our interval.
Step 4:
The interval of convergence is  ${\displaystyle (-1,1].}$

(a)     The radius of convergence is  ${\displaystyle R=1.}$
(b)     ${\displaystyle (-1,1]}$