Q12P - Mathematical Methods in Physical Sciences

Question

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

 $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{n}\mathbf{!}\mathbf{)}}^{\mathbf{2}}{\mathbf{z}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Step-by-Step Solution

Answer

Hence, the required disk of convergence is .

 $\left|z\right|<4$

Step 1: Define 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: $\sum _{n=0}^{\infty }\frac{{(n!)}^2{z}^n}{\left(2n\right)!}$, where, .$a_n=\frac{{(n!)}^2{z}^n}{\left(2n\right)!}$

Now, let us evaluate the ratio as: 

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

On further evaluating as:

 $\begin{array}{c}\rho =\underset{n\to \infty }{\lim }|\frac{z}{2}\cdot \frac{{(n+1)}^2}{(n+1)(2n+1)}|\\ =\frac{\left|z\right|}{2}\underset{n\to \infty }{\lim }\left|\frac{(n+1)}{(2n+1)}\right|\\ =\frac{\left|z\right|}{2}\underset{n\to \infty }{\lim }\left|\frac{(1+\frac{1}{n})}{(2+\frac{1}{n})}\right|\\ =\frac{\left|z\right|}{2}\left|\frac{(1+0)}{(2+0)}\right|=\frac{\left|z\right|}{4}\end{array}$

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

 $\begin{array}{c}\rho <1\\ \frac{\left|z\right|}{4}<1\\ \left|z\right|<4\end{array}$

Hence, the required disk of convergence is.$\left|z\right|<4$