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

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

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

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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {{\big (}{\frac {1}{2n+1}}{\big )}}{{\big (}{\frac {1}{n}}{\big )}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n}{2n+1}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$
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  $(2,4].$

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 of convergence is  $(2,4].$

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