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

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.

Final Answer:
(a)     The radius of convergence is  $R=1.$
(b)

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