009C Sample Final 3, Problem 6

Consider the power series

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n+1}}{n+1}}}$

(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  ${\displaystyle f(x)}$  to which the power series converges.

(d) Does the series

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

converge?

Foundations:
1. 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.

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

Solution:

(a)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x)^{n+2}}{(n+2)}}{\frac {n+1}{(-1)^{n}(x)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x){\frac {n+1}{n+2}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x|{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}}$

Step 2:
The Ratio Test tells us this series is absolutely convergent if  ${\displaystyle |x|<1.}$
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  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n+1}}.}$
This is an alternating series.
Let  ${\displaystyle b_{n}={\frac {1}{n+1}}.}$.
First, we have
${\displaystyle {\frac {1}{n+1}}\geq 0}$
for all  ${\displaystyle n\geq 0.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{n+2}}<{\frac {1}{n+1}}}$
for all  ${\displaystyle n\geq 0.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n+1}}=0.}$
Therefore, this series converges by the Alternating Series Test
and we include  ${\displaystyle x=1}$  in our interval.
Step 3:
Now, let  ${\displaystyle x=-1.}$
Then, the series becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{2n+1}}{n+1}}}&=&\displaystyle {\sum _{n=1}^{\infty }{\frac {-1}{n+1}}}\\&&\\&=&\displaystyle {(-1)\sum _{n=1}^{\infty }{\frac {1}{n+1}}.}\end{array}}}$
Now, we note that
${\displaystyle {\frac {1}{n+1}}>0}$
for all  ${\displaystyle n\geq 0.}$
This means that we can use the limit comparison test on this series.
Let  ${\displaystyle a_{n}={\frac {1}{n+1}}.}$
Let  ${\displaystyle b_{n}={\frac {1}{n}}.}$
Then,  ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$  diverges since it is the harmonic series.
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {({\frac {1}{n+1}})}{({\frac {1}{n}})}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{n+1}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }1.}\end{array}}}$
Therefore, the series
${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n+1}}}$
diverges by the Limit Comparison Test.
Therefore, we do not include  ${\displaystyle x=-1}$  in our interval.
Step 4:
The interval of convergence is  ${\displaystyle (-1,1].}$

(c)

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

(d)

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

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