# 009C Sample Final 2, Problem 4

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

$\sum _{n=1}^{\infty }(-1)^{n}{\frac {x^{n}}{n}}.$ (b) Find the interval of convergence of the above series.

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

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x)^{n+1}}{(n+1)}}\cdot {\frac {n}{(-1)^{n}(x)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x){\frac {n}{n+1}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x|{\frac {n}{n+1}}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n}{n+1}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}$ Step 2:
The Ratio Test tells us this series is absolutely convergent if  $|x|<1.$ Hence, the Radius of Convergence of this series is  $R=1.$ (b)

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