009C Sample Final 1, Problem 5

Let

$f(x)=\sum _{n=1}^{\infty }nx^{n}$

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

(b) Determine the interval of convergence of the power series.

(c) Obtain an explicit formula for the function $f(x)$.

ExpandFoundations:
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. After you find the radius of convergence, you need to check the endpoints of your interval
for convergence since the Ratio Test is inconclusive when $L=1.$

Solution:

(a)

ExpandStep 1:
To find the radius of convergence, we use the ratio test. We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)x^{n+1}}{nx^{n}}}}{\bigg |}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {n+1}{n}}x{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n}}}\\&&\\&=&\displaystyle {|x|.}\\\end{array}}$

ExpandStep 2:
Thus, we have $|x|<1$ and the radius of convergence of this series is $1.$

(b)

ExpandStep 1:
From part (a), we know the series converges inside the interval $(-1,1).$
Now, we need to check the endpoints of the interval for convergence.

ExpandStep 2:
For $x=1,$ the series becomes $\sum _{n=1}^{\infty }n,$ which diverges by the Divergence Test.

ExpandStep 3:
For $x=-1,$ the series becomes $\sum _{n=1}^{\infty }(-1)^{n}n,$ which diverges by the Divergence Test.
Thus, the interval of convergence is $(-1,1).$

(c)

ExpandStep 1:
Recall that we have the geometric series formula ${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}$ for $|x|<1.$
Now, we take the derivative of both sides of the last equation to get
${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}.$

ExpandStep 2:
Now, we multiply the last equation in Step 1 by $x.$
So, we have
${\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}=f(x).$
Thus, $f(x)={\frac {x}{(1-x)^{2}}}.$

ExpandFinal Answer:
(a)     $1$
(b)     $(-1,1)$
(c)     $f(x)={\frac {x}{(1-x)^{2}}}$

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