009C Sample Final 3, Problem 6

Consider the power series

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

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  $f(x)$  to which the power series converges.

(d) Does the series

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

converge? If so, find its sum.

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. Direct Comparison Test
Let  $\{a_{n}\}$  and  $\{b_{n}\}$  be positive sequences where  $a_{n}\leq b_{n}$
for all  $n\geq N$  for some  $N\geq 1.$
1. If  $\sum _{n=1}^{\infty }b_{n}$  converges, then  $\sum _{n=1}^{\infty }a_{n}$  converges.
2. If  $\sum _{n=1}^{\infty }a_{n}$  diverges, then  $\sum _{n=1}^{\infty }b_{n}$  diverges.

Solution:

(a)

Step 1:
We use the Ratio Test to determine the radius of convergence.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(-1)^{n+1}(x)^{n+2}}{(n+2)}}{\frac {n+1}{(-1)^{n}(x)^{n+1}}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}(-1)(x){\frac {n+1}{n+2}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }|x|{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n+2}}}\\&&\\&=&\displaystyle {|x|.}\end{array}}$

Step 2:
The Ratio Test tells us this series is absolutely convergent if  $|x|<1.$
Hence, the Radius of Convergence of this series is  $R=1.$

(b)

Step 1:
First, note that  $|x|<1$  corresponds to the interval  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).}
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 (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.}

Step 2:
First, let  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 x=1.}
Then, the series becomes  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 \sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.}
This is an alternating series.
Let  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 b_n=\frac{1}{n+1}.} .
The sequence  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 \{b_n\}}   is decreasing since
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 \frac{1}{n+2}<\frac{1}{n+1}}
for all  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 n\ge 0.}
Also,
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 \lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n+1}=0.}
Therefore, this series converges by the Alternating Series Test
and we include  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 x=1}   in our interval.

Step 3:
Now, let  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 x=-1.}
Then, the series becomes
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 \begin{array}{rcl} \displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\ &&\\ & = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.} \end{array}}
Now, we note that
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 \frac{1}{n+1}>0}
for all  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 n\ge 0.}
This means that we can use the limit comparison test on this series.
Let  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 a_n=\frac{1}{n+1}.}
Let  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 b_n=\frac{1}{n}.}
Then,  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 \sum_{n=1}^\infty b_n}   diverges since it is the harmonic series.
We have
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 \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} 1.} \end{array}}
Therefore, the series
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 \sum_{n=1}^{\infty} \frac{1}{n+1}}
diverges by the Limit Comparison Test.
Therefore, we do not include  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 x=-1}   in our interval.

Step 4:
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].}

(c)

Step 1:

Step 2:

(d)

Step 1:

Step 2:

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.}
(b)     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]}
(c)
(d)

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