009C Sample Midterm 3, Problem 5 Detailed Solution

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Find the radius of convergence and the interval of convergence of the series.

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

(b)  ${\displaystyle {\displaystyle \sum _{n=0}^{\infty }}{\frac {(x+1)^{n}}{n^{2}}}}$

Background Information:
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.

Solution:

(a)

Step 1:
We first use the Ratio Test to determine the radius of convergence.
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}x^{n+1}}{n+1+1}}{\frac {n+1}{(-1)^{n}x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)x{\frac {n+1}{n+2}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n+1}{n+2}}|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|(1)}\\&&\\&=&\displaystyle {|x|.}\end{array}}}$
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  ${\displaystyle R=1.}$
Step 3:
Now, we need to determine the interval of convergence.
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  ${\displaystyle L=1.}$
Step 4:
First, let  ${\displaystyle x=-1.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n+1}}.}$
We note that  ${\displaystyle {\frac {1}{n+1}}>0}$  for all  ${\displaystyle n\geq 0.}$
Hence, we can use the limit comparison test for this series.
Let  ${\displaystyle a_{n}={\frac {1}{n+1}}}$  and  ${\displaystyle b_{n}={\frac {1}{n}}.}$
Notice that  ${\displaystyle b_{n}>0}$  for all  ${\displaystyle n\geq 1.}$
The series  ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}$  is a  ${\displaystyle p}$-series with  ${\displaystyle p=1.}$  So,  ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$  diverges.
Now,

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {\frac {1}{n+1}}{\frac {1}{n}}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{n+1}}}\\&&\\&=&\displaystyle {1.}\end{array}}}$

Therefore, the series  ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n+1}}}$   diverges by the limit comparison test.
Hence, we do not include  ${\displaystyle x=-1}$  in the interval.
Step 5:
Now, 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
${\displaystyle {\frac {1}{(n+1)+1}}<{\frac {1}{n+1}}}$
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 6:
The interval of convergence is  ${\displaystyle (-1,1].}$

(b)

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

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

Step 2:
The Ratio Test tells us this series is absolutely convergent if  ${\displaystyle |x+1|<1.}$
Hence, the Radius of Convergence of this series is  ${\displaystyle R=1.}$
Step 3:
Now, we need to determine the interval of convergence.
First, note that  ${\displaystyle |x+1|<1}$  corresponds to the interval  ${\displaystyle (-2,0).}$
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 4:
First, let  ${\displaystyle x=-2.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n^{2}}}.}$
This is an alternating series.
Let  ${\displaystyle b_{n}={\frac {1}{n^{2}}}.}$.
First, we have
${\displaystyle {\frac {1}{n^{2}}}\geq 0}$
for all  ${\displaystyle n\geq 1.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{(n+1)^{2}}}<{\frac {1}{n^{2}}}}$
for all  ${\displaystyle n\geq 1.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n^{2}}}=0.}$
Therefore, this series converges by the Alternating Series Test
and we include  ${\displaystyle x=-2}$  in our interval.
Step 5:
Now, let  ${\displaystyle x=0.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}}}.}$
This is a  ${\displaystyle p}$-series with  ${\displaystyle p=2.}$
Hence, this series converges.
Therefore, we do include  ${\displaystyle x=0}$  in our interval.
Step 6:
The interval of convergence is  ${\displaystyle [-2,0].}$

(a)     The radius of convergence is  ${\displaystyle R=1}$  and the interval of convergence is  ${\displaystyle (-1,1].}$
(b)     The radius of convergence is  ${\displaystyle R=1}$  and the interval of convergence is  ${\displaystyle [-2,0].}$