Find the Radius of Convergence of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)(2n-1)!}$

Question

Find radius of convergence of series: $$ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)(2n-1)!} $$ I found out a hint to split my series into two: $$ b_n = \begin{cases} a_{2n-1} & \text{if $n$ is odd} \\ 0 &\text{else}\end{cases} $$ The radius of even $b_n$ is 0 but there is limit superior in ratio test. And since there is an absolute value when calculating the radius, $(-1)^n$ will be irrelevant.

I can then treat my odd-only series as: $$ \sum_{n=1}^{\infty}\frac{x^n}{n\cdot n!} $$ Leaving me with: $$ \varlimsup_{n\rightarrow+\infty} \left| \frac{b_{n+1}}{b_{n}} \right| = \varlimsup_{n\rightarrow+\infty} \frac{n}{n \cdot (n+1)} = 0 $$ And that means the radius of convergence is $\mathbb{R}$. Is that a correct solution?

Answer 1

It looks correct, but, instead of splitting the series, it is much more natural to do\begin{align}\lim_{n\to\infty}\frac{\left\lvert(-1)^{n+2}\frac{x^{2n+1}}{(2n+1)(2n+1)!}\right\rvert}{\left\lvert(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)(2n-1)!}\right\rvert}&=\lim_{n\to\infty}\lvert x\rvert^2\frac{2n-1}{2n(2n+1)^2}\\&=0.\end{align}So, your series converges for every $x$, which means that the radius of convergence is $\infty$.

Answer 2

Use Cauchy-Hadamard directly to obtain the convergence radius $\;R\;$:

$$\frac1R=\lim\sup_{n\to\infty}\sqrt[n]{b_n|}=\lim_{n\to\infty}\frac1{\sqrt[n]{(2n-1)(2n-1)!}}=0\implies R=\infty$$