How to find Radii of Convergence for power series with alternating 0s

Question

I was working on finding the Radius of Convergence of $\cosh{x}$, which is $1 + \frac{x^2}{2}+ \frac{x^4}{24}+ \frac{x^6}{720}+ \frac{x^8}{8!} + \cdots$.

The radius of convergence can be found with $\lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right|$, but because there are ALTERNATING 0's, this evaluates to 0 if n is odd, but $\infty$ when n is even...

For cosh x the radius is indeed infinity, but how can we use the above expression for the radius of convergence for a general power series with alternating 0 coefficients?

Answer 1

Your series is $g(x)=f(x^2)$ with $$f(y)=\sum a_ny^n,\quad a_n=\frac1{(2n)!}.$$ The radius of convergence of $f$ is $R=\lim\frac{a_n}{a_{n+1}}=+\infty$, hence the radius of convergence of $g$ is $\sqrt R=+\infty$.

Answer 2

Another way is to apply the d'Alembert test to the numerical series. Then for $x\neq 0$ we get $${x^{2n+2}\over (2n+2)!}{(2n)!\over x^{2n}}={x^2\over (2n+2)(2n+1)}\to 0$$ Hence the series is convergent for all $x,$ i.e. $R=\infty.$

The same method can be used to, for example, $\sum {x^{n^2}\over n!},$ where the method used in the comment by @geetha290krm is not available.