Q11P - Mathematical Methods in Physical Sciences

Question

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

 $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{n}\mathbf{!}\mathbf{)}}^{\mathbf{3}}{\mathbf{z}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{3}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

 

Step-by-Step Solution

Answer

Hence, the required disk of convergence is.$\left|z\right|<27$

Step 1: Disk of Convergence

For any power series $\sum a_n{z}^n$where z   is a complex numbers, then disk of convergence is given by: .$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{\left|}\frac{\mathbf{z}\mathbf{\times }\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}$

Step 2:Find the disk of Convergence

The given power series is:$\sum _{n=0}^{\infty }\frac{{(n!)}^3{z}^n}{\left(3n\right)!}$ , where$a_n=\frac{{(n!)}^3{z}^n}{\left(3n\right)!}$

Now, let us evaluate the ratio as: 

 $\begin{array}{c}\rho =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\frac{{((n+1)!)}^3{z}^{n+1}}{\left(3(n+1)\right)!}}{\frac{{(n!)}^3{z}^n}{\left(3n\right)!}}\right|\\ =\underset{n\to \infty }{\lim }|z\cdot \frac{{(n+1)}^3{(n!)}^3}{{(n!)}^3}\cdot \frac{\left(3n\right)!}{3(n+1)(3n+2)(3n+1)\left(3n\right)!}|\\ =\underset{n\to \infty }{\lim }|\frac{z}{3}\cdot \frac{{(n+1)}^3}{(n+1)(3n+2)(3n+1)}|\end{array}$

 

On further evaluating as:

$\begin{array}{c}\rho =\underset{n\to \infty }{\lim }|\frac{z}{3}\cdot \frac{{(n+1)}^3}{(n+1)(3n+2)(3n+1)}|<1\\ =\frac{\left|z\right|}{3}\underset{n\to \infty }{\lim }\left|\frac{{(n+1)}^2}{(3n+2)(3n+1)}\right|\\ =\frac{\left|z\right|}{3}\underset{n\to \infty }{\lim }\left|\frac{{(1+\frac{1}{n})}^2}{(3+\frac{2}{n})(3+\frac{1}{n})}\right|\\ =\frac{\left|z\right|}{3}\left|\frac{{(1+0)}^2}{(3+0)(3+0)}\right|=\frac{\left|z\right|}{27}\end{array}$

Now, for the series to be convergent, we have$\rho <1$  . So,

 $\begin{array}{c}\rho <1\\ \frac{\left|z\right|}{27}<1\\ \left|z\right|<27\end{array}$

Hence, the required disk of convergence is.$\left|z\right|<27$