# 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 $f(x)$ .
Foundations:
Recall:
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)

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

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

Step 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
${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}.$ Step 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, $f(x)={\frac {x}{(1-x)^{2}}}.$ (a) $1$ (b) $(-1,1)$ (c) $f(x)={\frac {x}{(1-x)^{2}}}$ 