<s>[INST] What is the radius of convergence of: What is the radius of convergence of the series
$$1+z+{z}^2/2^2+z^3/3!+z^4/2^4+z^5/5!+...$$
I used ratio test but then it gives two values of $R$
the values of $R$ are 0 and $\sqrt2$,
 [/INST] $$1 + z + \frac{z^2}{2^2} + \frac{z^3}{3!} + \frac{z^4}{2^4} + \cdots
   = \left(1 + \frac{z^2}{2^2} + \frac{z^4}{2^4} + \cdots\right) + \left(z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots \right)$$
We can rearrange terms like this as long as the series is absolutely convergent.  So let's investigate that.  First note that
$$ \left(1 + \frac{z^2}{2^2} + \frac{z^4}{2^4} + \cdots\right) = \sum_{n=0}^{+\infty} \left(\frac{z}{2}\right)^{2n}$$
and
$$ \left(z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots \right) = \sum_{n=0}^{+\infty} \frac{z^{2n+1}}{(2n+1)!}$$
Let $R_1$ be the radius of convergence of the first series and $R_2$ be the radius of convergence of the second series.
Apply the ratio test to the first series to get $R_1 = 2$.  Note that if $|z| = 2$ then the first series will diverge.  So the disk of convergence for the first series is all $z \in \mathbb{C}$ such that $|z| < 2$.  Apply the ratio test to the second series to get $R_2 = +\infty$.  So the second series converges on all of $\mathbb{C}$.
We need both of these series to be absolutely convergent in order for the original series to converge.  Therefore the original series will converge on the intersection of the domains of convergence for the other two series.  Since the second series converges on all of $\mathbb{C}$ and the first series converges on $\{z \in \mathbb{C} : |z| < 2\}$, then the original series converges on $\{z \in \mathbb{C} : |z| < 2\}$.  Therefore the radius of convergence of the original series is 2.
EDIT - Verifying $R_1$:
$a_n = \dfrac{z^{2n}}{2^{2n}} = \dfrac{z^{2n}}{4^n}$.  Therefore $a_{n+1} = \dfrac{z^{2n+2}}{4^{n+1}} = \dfrac{z^{2n}z^2}{4^n4}.$
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{z^{2n}z^2}{4^n4} \cdot \frac{4^n}{z^{2n}}\right| = \frac{|z|^2}{4}$$
Take the limit as $n \to +\infty$, still get $|z|^2 / 4$.  Solve $|z|^2 / 4 < 1$ to get $|z| < 2$.
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