Question

Find the disk of convergence for each of the following complex power series.

$\mathit{z}\mathbf{-}\frac{{\mathbf{z}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{z}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{\dots }$

Short answer

The disc of convergence is $\left|z\right|<1$.

Step-by-step solution

Given Information.

The given power series, i.e., ${\mathrm{S}}_{\mathrm{n}}=\mathrm{z}-\frac{{\mathrm{Z}}^2}{2}+\frac{{\mathrm{Z}}^3}{3}-\frac{{\mathrm{z}}^4}{4}+\dots$

Definition of Disc of convergence.

The interior of the set of points of convergence of a power series is called the disc of convergence. Its radius is known as the series' convergence radius.

Find the general term of the series.

Use the series to find general terms.

$$\begin{gathered} \begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\mathrm{z}-\frac{{\mathrm{z}}^2}{2}+\frac{{\mathrm{z}}^3}{3}-\frac{{\mathrm{z}}^4}{4}+......\left(1\right)\end{array} \\ {\mathrm{S}}_{\mathrm{n}}=\underset{}{{\displaystyle \sum _{\mathrm{n}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}+1}{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}}}...\left(2\right) \end{gathered}$$

Find the disc of convergence.

Use the ratio test.

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ {\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\displaystyle \frac{{\left(-1\right)}^{\mathrm{n}+2}{\mathrm{z}}^{\mathrm{n}+1}}{\mathrm{n}+1}}}{{\displaystyle \frac{{\left(-1\right)}^{\mathrm{n}+1}{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}}}}\right| \\ =\left|\left(-1\right)\mathrm{z}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)\right| \\ =\left|\left(-1\right)\mathrm{z}\left(\frac{1}{1+1/\mathrm{n}}\right)\right| \end{gathered}$$

Calculate the value of $\mathrm{\rho }$, i.e.,

role="math" localid="1658731500708" $$\begin{gathered} \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{\rho }}_{\mathrm{n}} \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\left(-1\right)\mathrm{z}\left(\frac{1}{1+{\displaystyle \frac{1}{\mathrm{n}}}}\right)\right| \\ =\left|\mathrm{z}\right| \end{gathered}$$

$S_n$is convergent for$\mathrm{\rho }<1$, i.e., $\left|\mathrm{z}\right|<1$.

Hence, the disc of convergence is $\left|\mathrm{z}\right|<1$.