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  $a_{n}<b_{n}$  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.

Final Answer:
(a)     The radius of convergence is  $R=1.$
(b)     $(-1,1]$
(c)     $f(x)=\ln(1+x)$
(d)     converges (by the Direct Comparison Test)

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