to find radius of convergence of power series.

Question

I have a power series given as:

$f(z) =1 + z+ \frac{z^2}{2^2} +\frac{z^3}{3!} + \frac{z^4}{2^4} \frac{z^2}{2^2}+ \frac{z^5}{5!}+ \ldots$

I have to find radius of convergence of above series. My attempt :

I divided the series into two series

$1+ \frac{z^2}{2^2} + \frac{z^4}{2^4}+ \ldots$ and

$ z+\frac{z^3}{3!} + \frac{z^5}{5!}+ \ldots$

For 1st series $a_n = \frac{1}{2^{2n}}$ and for 2nd series it is $\frac{1}{n!}$.

After solving ROC of 1st series comes out to be 2 and of that other series it is 1. But my answer is wrong. Where I went wrong? For 1st series I used the formula $R = \frac{a_n}{a_{n+1}}$ with limit n tends to $\infty$ and replaced $z^2$ by $s$.

Answer 1

I think for this one it is much easier to use logic.

You have divided the sum into two parts which are manageable and known, which is good. The first is

$$\sum_{n=0}^{\infty} \left(\frac{z^2}{4}\right)^n$$

Which converges when $|z|<2$ (the magnitude of the inside must be less than one).

The other sum you got looks a lot like $e^z$. In fact, it is less than $e^z$ as $z\rightarrow \infty$, so by a quick comparison test it has an infinite radius of convergence.

Therefore, since the first sum diverges when $|z|<2$, so does the original sum from the problem.

Answer 2

Hint:

$$f(z)=\frac4{4-z^2}+\frac{e^z-e^{-z}}2$$