Question

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

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

Short answer

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

Step-by-step solution

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

Now, let us evaluate the ratio as:

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

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

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

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