A function $f$ is defined by the power series $f(x)=\sum\limits_{k\,=\,0}^{\infty }x^{k}$.

Solution (a) $\sum\limits_{k\,=\,0}^{\infty }x^{k}$ is a power series centered at $0$ with $a_{k}=1$. Then $$f(x) =1+x+x^{2}+x^{3}+x^{4}+ \cdots$$

The domain of $f$ equals the interval of convergence of the power series. Since the series $\sum\limits_{k\,=\,0}^{\infty }x^{k}$ is a geometric series, it converges for $\vert x\vert < 1$. The radius of convergence is $1$, and the interval of convergence is $(-1,1)$. The domain of $f$ is the open interval $(-1,1)$.

605

(b) The numbers $\dfrac{1}{2}$ and $-\dfrac{1}{3}$ are in the interval $(-1,1)$, so they are in the domain of $f$. Then $f\left( \dfrac{1}{2}\right)$ is a geometric series with $r=\dfrac{1}{2}$, $a=1$, and $$f\left(\dfrac{1}{2}\right) =1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right) ^{2}+\left(\dfrac{1}{2}\right) ^{3}+\cdots \,= \frac{a}{1-r} = \dfrac{1}{1-\dfrac{1}{2}}=2$$

Similarly, $$f\left( -\dfrac{1}{3}\right) =1-\dfrac{1}{3}+\left( -\dfrac{1}{3}\right) ^{2}+\left( -\dfrac{1}{3}\right) ^{3}+\cdots \,=\dfrac{1}{1+\dfrac{1}{3}}=\dfrac{3}{4}$$

(c) Since $f$ is defined by a geometric series, we can find $f$ by summing the series. $$\begin{eqnarray*} &&f(x) =\sum\limits_{k\,=\,0}^{\infty }x^{k} = 1 + x + x^{2} + \cdots + x^{n} + \cdots \underset{\underset{\color{#00ADEE}{a=1; r = x }}{\color{#00ADEE}{\uparrow}}}{=}\dfrac{1}{1-x}\quad -1 < x < 1\\ \end{eqnarray*}$$