009C Sample Midterm 2, Problem 4

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

(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty n^nx^n}

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

Foundations:
1. Root Test
Let  $\{a_{n}\}$  be a positive sequence and let  $\lim _{n\rightarrow \infty }|a_{n}|^{\frac {1}{n}}=L.$
Then,
If  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|a_{n}|}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt[{n}]{|n^{n}x^{n}|}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|n^{n}x^{n}|^{\frac {1}{n}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|nx|}\\&&\\&=&\displaystyle {n|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }n}\\&&\\&=&\displaystyle {\infty .}\end{array}}$

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  $[-2,0).$

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 of convergence is  $[-2,0).$

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