# 009C Sample Midterm 2, Problem 4 Detailed Solution

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

(a)  $\sum _{n=1}^{\infty }n^{n}x^{n}$ (b)  $\sum _{n=1}^{\infty }{\frac {(x+1)^{n}}{\sqrt {n}}}$ Background Information:
1. Root Test
Let  $\{a_{n}\}$ be a positive sequence and let  $\lim _{n\rightarrow \infty }|a_{n}|^{\frac {1}{n}}=L.$ Then,
If  $L<1,$ the series is absolutely convergent.

If  $L>1,$ the series is divergent.

If  $L=1,$ the test is inconclusive.

2. 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 begin by applying the Root Test.
We have

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|n^{n}x^{n}|}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|n^{n}x^{n}|^{\frac {1}{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|nx|}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }n}\\&&\\&=&\displaystyle {\infty .}\end{array}}$ Step 2:
This means that as long as  $x\neq 0,$ this series diverges.
Hence, the radius of convergence is  $R=0$ and
the interval of convergence is  $\{0\}.$ (b)

Step 1:
We first use the Ratio Test to determine the radius of convergence.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(x+1)^{n+1}}{\sqrt {n+1}}}{\frac {\sqrt {n}}{(x+1)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(x+1){\frac {\sqrt {n}}{\sqrt {n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x+1|{\frac {\sqrt {n}}{\sqrt {n+1}}}}\\&&\\&=&\displaystyle {|x+1|\lim _{n\rightarrow \infty }{\sqrt {\frac {n}{n+1}}}}\\&&\\&=&\displaystyle {|x+1|{\sqrt {\lim _{n\rightarrow \infty }{\frac {n}{n+1}}}}}\\&&\\&=&\displaystyle {|x+1|{\sqrt {1}}}\\&&\\&=&\displaystyle {|x+1|.}\end{array}}$ 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  $x=0.$ Then, the series becomes  $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}.$ We note that this is a  $p$ -series with  $p={\frac {1}{2}}.$ Since  $p<1,$ the series diverges.
Hence, we do not include  $x=0$ in the interval.
Step 5:
Now, let  $x=-2.$ Then, the series becomes  $\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{\sqrt {n}}}.$ This series is alternating.
Let  $b_{n}={\frac {1}{\sqrt {n}}}.$ First, we have
${\frac {1}{\sqrt {n}}}\geq 0$ for all  $n\geq 1.$ The sequence  $\{b_{n}\}$ is decreasing since
${\frac {1}{\sqrt {n+1}}}<{\frac {1}{\sqrt {n}}}$ for all  $n\geq 1.$ Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{\sqrt {n}}}=0.$ Therefore, the series converges by the Alternating Series Test.
Hence, we include  $x=-2$ in our interval of convergence.
Step 6:
The interval of convergence is  $[-2,0).$ (a)     The radius of convergence is  $R=0$ and the interval of convergence is  $\{0\}.$ (b)     The radius of convergence is  $R=1$ and the interval of convergence is  $[-2,0).$ 