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:
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  $n$th term test.

Step 3:
For  $x=-1,$   the series becomes  $\sum _{n=1}^{\infty }(-1)^{n}n,$  which diverges by the  $n$th term 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,  $f(x)={\frac {x}{(1-x)^{2}}}.$

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

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