Q6P - Mathematical Methods in Physical Sciences

Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}{\mathit{n}}^{\mathbf{2}}{\mathbf{\left(}\mathbf{3}\mathit{i}\mathit{z}\mathbf{\right)}}^{\mathbf{n}}\mathbf{.}$ 

Step-by-Step Solution

Answer

The region of convergence is $\left|z\right|<\frac{1}{3}.|$ .

Step 1: Given Information.

The given power series, i.e.,  $S_n=\sum _{n-1}^{\infty }{n}^2{\left(3iz\right)}^n$

Step 2: Definition of Region of convergence.

The disc of convergence can be defined as the interior of the set of terms/points of the any converging series.

Step 3: Find the Region of convergence.

Use the ratio test. 

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ {\mathrm{\rho }}_{\mathrm{n}}=\frac{{(\mathrm{n}+1)}^2{\left(3\mathrm{iz}\right)}^{\mathrm{n}+1}}{{\mathrm{n}}^2{\left(3\mathrm{iz}\right)}^{\mathrm{n}}} \end{gathered}$$ 

 

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

$$\begin{gathered} \rho =\underset{n\to \infty }{\lim }\left|\frac{{(n+1)}^2{\left(3iz\right)}^{n+1}}{n^2{\left(3iz\right)}^n}\right| \\ =\underset{n\to \infty }{\lim }\left|{\left(\frac{n+1}{n}\right)}^2\left(3iz\right)\right| \\ =\underset{n\to \infty }{\lim }\left|{\left(1+\frac{1}{n}\right)}^2\left(3iz\right)\right| \\ =\left|3iz\right| \end{gathered}$$ 

 

 $S_n$ is convergent for $\mathrm{\rho }<1$ , i.e., $\left|3iz\right|<1$ .

 

Rewrite $\left|3iz\right|<1$ as