Find the radius of convergence of the Power series $1+z+\frac{z^2}{2^2}+\frac{z^3}{3!}+\frac{z^4}{2^4}+\dots$

Question

Find the radius of convergence of the Power series $$1+z+\frac{z^2}{2^2}+\frac{z^3}{3!}+\frac{z^4}{2^4}+\frac{z^5}{5!}\cdots $$

Put the series in the form

$$\left[1+\frac{z^2}{2^2}+\frac{z^4}{2^4}+\cdots\right]+\left[z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right] =\left[\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}\right] +\sinh(z)$$ The radius of convergence of the $1$st series is $4$ and, since $-\infty<\sinh(z)<\infty$, the radius of convergence of the given series is $4$.

Is my approach right?

Answer 1

Your approach is in the right direction.

The first series $\displaystyle\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}$ has radius of convergence 2 - try the root test for example, while the second one has infinite radius of convergence. Thus their sum has radius of convergence equal to 2.

Root test for $\{a_n\}_{n\in\mathbb N}$, where $$ a_n=\left\{\begin{array}{lll} 2^{-n} & \text{if} & \text{$n$ even}, \\ 0 & \text{if} & \text{$n$ odd}, \end{array}\right. $$ is $$ \limsup \lvert a_n\rvert^{1/n}=\frac{1}{2}, $$ and hence radius of convergence$=2$.

Note that $\displaystyle\sum_{n=0}^\infty\frac{z^{2n}}{2^{2n}}=\sum_{n=0}^\infty a_nz^n$.