# 009C Sample Midterm 1, Problem 5 Detailed Solution

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

(a)  $\sum _{n=0}^{\infty }{\sqrt {n}}x^{n}$ (b)  $\sum _{n=0}^{\infty }(-1)^{n}{\frac {(x-3)^{n}}{2n+1}}$ Background Information:
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 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  $b_{n}={\frac {1}{2n+1}}.$ .
First, we have
${\frac {1}{2n+1}}\geq 0$ for all  $n\geq 0.$ The sequence  $\{b_{n}\}$ is decreasing since
${\frac {1}{2(n+1)+1}}<{\frac {1}{2n+1}}$ for all  $n\geq 0.$ 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,
$\sum _{n=0}^{\infty }{\frac {1}{2n+1}}$ diverges by the Limit Comparison Test.
Therefore, we do not include  $x=2$ in our interval.
Step 6:
The interval of convergence is  $(2,4].$ (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 of convergence is  $(2,4].$ 