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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  $|x|<1.$
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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |x|<1}   corresponds to the interval  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=1.}

Step 4:
First, let  $x=1.$
Then, the series becomes  $\sum _{n=0}^{\infty }{\sqrt {n}}.$
We note that
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\sqrt {n}}.}
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }{\sqrt {n}}=\infty ,}
we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (-1,1).}

(b)

Step 1:
We first use the Ratio Test to determine the radius of convergence.
We have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=4.}
Then, the series becomes  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{2n+1}}.}
This is an alternating series.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b_{n}={\frac {1}{2n+1}}.} .
First, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{2n+1}}\geq 0}
for all  $n\geq 0.$
The sequence  $\{b_{n}\}$  is decreasing since
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{2(n+1)+1}}<{\frac {1}{2n+1}}}
for all  $n\geq 0.$
Also,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  $x=4$  in our interval.

Step 5:
Now, let  $x=2.$
Then, the series becomes  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{2n+1}}.}
First, we note that  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}},}
which is the harmonic series and divergent.
Now, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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,
$\sum _{n=0}^{\infty }{\frac {1}{2n+1}}$
diverges by the Limit Comparison Test.
Therefore, we do not include  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=2}   in our interval.

Step 6:
The interval of convergence is  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (2,4].}

Final Answer:
(a)     The radius of convergence is  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 R=1}   and the interval of convergence is  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 (-1,1).}
(b)     The radius of convergence is  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 R=1}   and the interval of convergence is  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 (2,4].}

Return to Sample Exam