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:
Review ratio test.

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, $f(x)={\frac {x}{(1-x)^{2}}}$.

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

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