# 009C Sample Midterm 1, Problem 5 Detailed Solution

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

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

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

Background Information:
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 first use the Ratio Test to determine the radius of convergence.
We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {{\sqrt {n+1}}x^{n+1}}{{\sqrt {n}}x^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {\sqrt {n+1}}{\sqrt {n}}}x{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\sqrt {\frac {n+1}{n}}}|x|}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\sqrt {\frac {n+1}{n}}}}\\&&\\&=&\displaystyle {|x|{\sqrt {\lim _{n\rightarrow \infty }{\frac {n+1}{n}}}}}\\&&\\&=&\displaystyle {|x|{\sqrt {1}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}}$
Step 2:
The Ratio Test tells us this series is absolutely convergent if  ${\displaystyle |x|<1.}$
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  ${\displaystyle |x|<1}$  corresponds to the interval  ${\displaystyle (-1,1).}$
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  ${\displaystyle L=1.}$
Step 4:
First, let  ${\displaystyle x=1.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }{\sqrt {n}}.}$
We note that
${\displaystyle \lim _{n\rightarrow \infty }{\sqrt {n}}=\infty .}$
Therefore, the series diverges by the  ${\displaystyle n}$th term test.
Hence, we do not include  ${\displaystyle x=1}$  in the interval.
Step 5:
Now, let  ${\displaystyle x=-1.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\sqrt {n}}.}$
Since  ${\displaystyle \lim _{n\rightarrow \infty }{\sqrt {n}}=\infty ,}$
we have
${\displaystyle \lim _{n\rightarrow \infty }(-1)^{n}{\sqrt {n}}={\text{DNE}}.}$
Therefore, the series diverges by the  ${\displaystyle n}$th term test.
Hence, we do not include  ${\displaystyle x=-1}$  in the interval.
Step 6:
The interval of convergence is  ${\displaystyle (-1,1).}$

(b)

Step 1:
We first use the Ratio Test to determine the radius of convergence.
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}}{\frac {2n+1}{(-1)^{n}(x-3)^{n}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x-3){\frac {2n+1}{2n+3}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x-3|{\frac {2n+1}{2n+3}}}\\&&\\&=&\displaystyle {|x-3|\lim _{n\rightarrow \infty }{\frac {2n+1}{2n+3}}}\\&&\\&=&\displaystyle {|x-3|.}\end{array}}}$

Step 2:
The Ratio Test tells us this series is absolutely convergent if  ${\displaystyle |x-3|<1.}$
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  ${\displaystyle |x-3|<1}$  corresponds to the interval  ${\displaystyle (2,4).}$
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  ${\displaystyle R=1.}$
Step 4:
First, let  ${\displaystyle x=4.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{2n+1}}.}$
This is an alternating series.
Let  ${\displaystyle b_{n}={\frac {1}{2n+1}}.}$.
First, we have
${\displaystyle {\frac {1}{2n+1}}\geq 0}$
for all  ${\displaystyle n\geq 0.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{2(n+1)+1}}<{\frac {1}{2n+1}}}$
for all  ${\displaystyle n\geq 0.}$
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{2n+1}}=0.}$
Therefore, this series converges by the Alternating Series Test
and we include  ${\displaystyle x=4}$  in our interval.
Step 5:
Now, let  ${\displaystyle x=2.}$
Then, the series becomes  ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{2n+1}}.}$
First, we note that  ${\displaystyle {\frac {1}{2n+1}}>0}$  for all  ${\displaystyle n\geq 0.}$
Thus, we can use the Limit Comparison Test.
We compare this series with the series  ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}},}$
which is the harmonic series and divergent.
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\frac {1}{2n+1}}{\frac {1}{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}}$

Since this limit is a finite number greater than zero,
${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{2n+1}}}$
diverges by the Limit Comparison Test.
Therefore, we do not include  ${\displaystyle x=2}$  in our interval.
Step 6:
The interval of convergence is  ${\displaystyle (2,4].}$

(a)     The radius of convergence is  ${\displaystyle R=1}$  and the interval of convergence is  ${\displaystyle (-1,1).}$
(b)     The radius of convergence is  ${\displaystyle R=1}$  and the interval of convergence is  ${\displaystyle (2,4].}$