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

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

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

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