Q7P - Mathematical Methods in Physical Sciences

Question

: Find the disk of convergence for each of the followingcomplex power series.

 $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{z}}^{\mathbf{2}\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Step-by-Step Solution

Answer

The required disk of convergence is .$\left|z\right|<\infty$

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

Now, let us evaluate the ratio as: 

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

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

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

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