009C Sample Midterm 2, Problem 4

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

a) $\sum _{n=0}^{\infty }n^{n}x^{n}$

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

Foundations:
Root Test
Ratio Test

Solution:

(a)

Step 1:
We begin by applying the Root Test.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\sqrt {|a_{n}|}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt {|n^{n}x^{n}|}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|n^{n}x^{n}|^{\frac {1}{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|nx|}\\&&\\&=&\displaystyle {n|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }n}\\&&\\&=&\displaystyle {\infty }\end{array}}$

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

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 $\sum _{n=0}^{\infty }{\frac {1}{\sqrt {n}}}.$
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 $\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{\sqrt {n}}}.$
This series is alternating.
Let $b_{n}={\frac {1}{\sqrt {n}}}.$
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 $[-2,0).$

Final Answer:
(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 fo convergence is $[-2,0).$

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