# 009C Sample Midterm 2, Problem 4 Detailed Solution

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

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

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

Background Information:
1. Root Test
Let  ${\displaystyle \{a_{n}\}}$  be a positive sequence and let  ${\displaystyle \lim _{n\rightarrow \infty }|a_{n}|^{\frac {1}{n}}=L.}$
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. 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 begin by applying the Root Test.
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|n^{n}x^{n}|}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|n^{n}x^{n}|^{\frac {1}{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|nx|}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }n|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }n}\\&&\\&=&\displaystyle {\infty .}\end{array}}}$

Step 2:
This means that as long as  ${\displaystyle x\neq 0,}$  this series diverges.
Hence, the radius of convergence is  ${\displaystyle R=0}$  and
the interval of convergence is  ${\displaystyle \{0\}.}$

(b)

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

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