Question on convergence of $\sum_{n=0}^{\infty} \frac{(-1)^n2^{-n}}{n!}z^{2n+1}$

Question

Let $z \in \mathbb{C}$. Consider the following power series:

$$ \sum_{n=0}^{\infty} \frac{(-1)^n2^{-n}}{n!}z^{2n+1}.$$

Find the convergence radius R of this series.

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My initial thoughts are to use the ratio test. However, I am not completely sure whether or not I need to change the index of the sum since one of the assumptions is the $a_n \neq0$ from a certain step and for even powers of $z$ $a_n$ is clearly 0. So can I use the test and if so what is the formal argument?

When using the ratio test the limit of the ratio test is $0$ $\implies$ the convergence radius $R=\infty$.

Answer 1

For $n \ge 0$ let $b_n:=\frac{(-1)^n2^{-n}}{n!}z^{2n+1}.$ Then compute , for $z \ne 0$,

$$ \lim_{n \to \infty } \frac{|b_{n+1}|}{|b_n|}.$$

You will see that your power series has radius of convergence $= \infty.$

Do you see a connection wit $exp$ ?

Answer 2

You need to rewrite the power series

$$z\sum_{n=0}^\infty\dfrac{(-1)^n2^{-n}}{n!}z^{2n}.$$

The radius of convergence depends on the infinite power series $$\sum_{n=0}^\infty\dfrac{(-1)^n2^{-n}}{n!}z^{2n}$$

with $z^2=x$ we can rewrite this sum as $$\sum_{n=0}^\infty\dfrac{(-1)^n2^{-n}}{n!}x^{n}.$$

The radius of convergence for this is $R_x=\infty$ which implies the radius of convergence $R_z = \infty$ for $z$.