# 009C Sample Final 3, Problem 6

Consider the power series

$\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n+1}}{n+1}}$ (a) Find the radius of convergence of the above power series.

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

(c) Find the closed formula for the function  $f(x)$ to which the power series converges.

(d) Does the series

$\sum _{n=0}^{\infty }{\frac {1}{(n+1)3^{n+1}}}$ converge?

Foundations:
1. 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.

2. Direct Comparison Test
Let  $\{a_{n}\}$ and  $\{b_{n}\}$ be positive sequences where  $a_{n}\leq b_{n}$ for all  $n\geq N$ for some  $N\geq 1.$ 1. If  $\sum _{n=1}^{\infty }b_{n}$ converges, then  $\sum _{n=1}^{\infty }a_{n}$ converges.
2. If  $\sum _{n=1}^{\infty }a_{n}$ diverges, then  $\sum _{n=1}^{\infty }b_{n}$ diverges.

Solution:

(a)

Step 1:
We use the Ratio Test to determine the radius of convergence.
We have

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x)^{n+2}}{(n+2)}}{\frac {n+1}{(-1)^{n}(x)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x){\frac {n+1}{n+2}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x|{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}$ 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  $\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n+1}}.$ This is an alternating series.
Let  $b_{n}={\frac {1}{n+1}}.$ .
First, we have
${\frac {1}{n+1}}\geq 0$ for all  $n\geq 0.$ The sequence  $\{b_{n}\}$ is decreasing since
${\frac {1}{n+2}}<{\frac {1}{n+1}}$ for all  $n\geq 0.$ Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n+1}}=0.$ 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
${\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{2n+1}}{n+1}}}&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {-1}{n+1}}}\\&&\\&=&\displaystyle {(-1)\sum _{n=1}^{\infty }{\frac {1}{n+1}}.}\end{array}}$ Now, we note that
${\frac {1}{n+1}}>0$ for all  $n\geq 0.$ This means that we can use the limit comparison test on this series.
Let  $a_{n}={\frac {1}{n+1}}.$ Let  $b_{n}={\frac {1}{n}}.$ Then,  $\sum _{n=1}^{\infty }b_{n}$ diverges since it is the harmonic series.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {({\frac {1}{n+1}})}{({\frac {1}{n}})}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{n+1}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }1.}\end{array}}$ Therefore, the series
$\sum _{n=1}^{\infty }{\frac {1}{n+1}}$ diverges by the Limit Comparison Test.
Therefore, we do not include  $x=-1$ in our interval.
Step 4:
The interval of convergence is  $(-1,1].$ (c)

Step 1:
Let
$f(x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n+1}}{n+1}}.$ Then,
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {d}{dx}}{\bigg (}\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n+1}}{n+1}}{\bigg )}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {d}{dx}}{\bigg (}(-1)^{n}{\frac {x^{n+1}}{n+1}}{\bigg )}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }(-1)^{n}x^{n}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }(-x)^{n}}\\&&\\&=&\displaystyle {\frac {1}{1-(-x)}}\\&&\\&=&\displaystyle {{\frac {1}{1+x}}.}\end{array}}$ Step 2:
Thus,
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\int {\frac {1}{1+x}}~dx}\\&&\\&=&\displaystyle {\ln(1+x)+C.}\end{array}}$ Since there is no constant term in the series $\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n+1}}{n+1}},$ $C=0.$ Hence,
$f(x)=\ln(1+x).$ (d)

Step 1:
First, we note that
${\frac {1}{(n+1)3^{n+1}}}>0$ for all  $n\geq 0.$ This means that we can use a comparison test on this series.
Let  $a_{n}={\frac {1}{(n+1)3^{n+1}}}.$ Step 2:
Let  $b_{n}={\frac {1}{3^{n+1}}}.$ We want to compare the series in this problem with
$\sum _{n=1}^{\infty }b_{n}=\sum _{n=1}^{\infty }{\frac {1}{3}}{\bigg (}{\frac {1}{3}}{\bigg )}^{n}.$ This is a geometric series with  $r={\frac {1}{3}}.$ Since   $|r|<1,$ the series  $\sum _{n=1}^{\infty }b_{n}$ converges.
Step 3:
Also, we have  $a_{n} since
${\frac {1}{(n+1)3^{n+1}}}<{\frac {1}{3^{n+1}}}$ for all  $n\geq 0.$ Therefore, the series  $\sum _{n=1}^{\infty }a_{n}$ converges
by the Direct Comparison Test.

(a)     The radius of convergence is  $R=1.$ (b)     $(-1,1]$ (c)     $f(x)=\ln(1+x)$ 