009C Sample Midterm 1, Problem 5

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

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

b) $\sum _{n=0}^{\infty }(-1)^{n}{\frac {(x-3)^{n}}{2n+1}}$

Foundations:
1. 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.
2. If a series absolutely converges, then it also converges.

Solution:

(a)

Step 1:
We first use the Ratio Test to determine the radius of convergence.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {{\sqrt {n+1}}x^{n+1}}{{\sqrt {n}}x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {\sqrt {n+1}}{\sqrt {n}}}x{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt {\frac {n+1}{n}}}|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\sqrt {\frac {n+1}{n}}}}\\&&\\&=&\displaystyle {|x|{\sqrt {\lim _{n\rightarrow \infty }{\frac {n+1}{n}}}}}\\&&\\&=&\displaystyle {|x|{\sqrt {1}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}$

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 $\sum _{n=0}^{\infty }{\sqrt {n}}.$
We note that
$\lim _{n\rightarrow \infty }{\sqrt {n}}=\infty .$
Therefore, the series diverges by the $n$th term test.
Hence, we do not include $x=1$ in the interval.

Step 5:
Now, let $x=-1.$
Then, the series becomes $\sum _{n=0}^{\infty }(-1)^{n}{\sqrt {n}}.$
Since $\lim _{n\rightarrow \infty }{\sqrt {n}}=\infty ,$
we have
$\lim _{n\rightarrow \infty }(-1)^{n}{\sqrt {n}}={\text{DNE}}.$
Therefore, the series diverges by the $n$th term test.
Hence, we do not include $x=-1$ in the 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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}}{\frac {2n+1}{(-1)^{n}(x-3)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x-3){\frac {2n+1}{2n+3}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x-3|{\frac {2n+1}{2n+3}}}\\&&\\&=&\displaystyle {|x-3|\lim _{n\rightarrow \infty }{\frac {2n+1}{2n+3}}}\\&&\\&=&\displaystyle {|x-3|.}\end{array}}$

Step 2:
The Ratio Test tells us this series is absolutely convergent if $|x-3|<1.$
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 $|x-3|<1$ corresponds to the interval $(2,4).$
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 $x=4.$
Then, the series becomes $\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{2n+1}}.$
This is an alternating series.
Let $b_{n}={\frac {1}{2n+1}}.$.
The sequence $\{b_{n}\}$ is decreasing since
${\frac {1}{2(n+1)+1}}<{\frac {1}{2n+1}}$
for all $n\geq 1.$
Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{2n+1}}=0.$
Therefore, this series converges by the Alternating Series Test
and we include $x=4$ in our interval.

Step 5:
Now, let $x=2.$
Then, the series becomes $\sum _{n=0}^{\infty }{\frac {1}{2n+1}}.$
First, we note that ${\frac {1}{2n+1}}>0$ for all $n\geq 0.$
Thus, we can use the Limit Comparison Test.
We compare this series with the series $\sum _{n=1}^{\infty }{\frac {1}{n}},$
which is the harmonic series and divergent.
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\frac {1}{2n+1}}{\frac {1}{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$
Since this limit is a finite number greater than zero, we have
$\sum _{n=0}^{\infty }{\frac {1}{2n+1}}$ diverges by the
Limit Comparison Test. Therefore, we do not include $x=2$
in our interval.

Step 6:
The interval of convergence is $(2,4].$

Final Answer:
(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 fo convergence is $(2,4].$

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