# 009C Sample Final 1, Problem 5

Let

${\displaystyle f(x)=\sum _{n=1}^{\infty }nx^{n}}$
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 ${\displaystyle f(x)}$.
Foundations:
Recall:
1. 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.
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 ${\displaystyle L=1.}$

Solution:

(a)

Step 1:
To find the radius of convergence, we use the ratio test. We have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)x^{n+1}}{nx^{n}}}}{\bigg |}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {n+1}{n}}x{\bigg |}}\\&&\\&=&\displaystyle {|x|\lim _{n\rightarrow \infty }{\frac {n+1}{n}}}\\&&\\&=&\displaystyle {|x|.}\\\end{array}}}$
Step 2:
Thus, we have ${\displaystyle |x|<1}$ and the radius of convergence of this series is ${\displaystyle 1.}$

(b)

Step 1:
From part (a), we know the series converges inside the interval ${\displaystyle (-1,1).}$
Now, we need to check the endpoints of the interval for convergence.
Step 2:
For ${\displaystyle x=1,}$ the series becomes ${\displaystyle \sum _{n=1}^{\infty }n,}$ which diverges by the Divergence Test.
Step 3:
For ${\displaystyle x=-1,}$ the series becomes ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}n,}$ which diverges by the Divergence Test.
Thus, the interval of convergence is ${\displaystyle (-1,1).}$

(c)

Step 1:
Recall that we have the geometric series formula ${\displaystyle {\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}}$ for ${\displaystyle |x|<1.}$
Now, we take the derivative of both sides of the last equation to get
${\displaystyle {\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}.}$
Step 2:
Now, we multiply the last equation in Step 1 by ${\displaystyle x.}$
So, we have
${\displaystyle {\frac {x}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n}=f(x).}$
Thus, ${\displaystyle f(x)={\frac {x}{(1-x)^{2}}}.}$
(a) ${\displaystyle 1}$
(b) ${\displaystyle (-1,1)}$
(c) ${\displaystyle f(x)={\frac {x}{(1-x)^{2}}}}$