I'm trying to show that $ \sum_{n=0}^{\infty} \dfrac{(-1)^n \, x^{2n+1}}{2^n\,n!}$ is absolutely convergent for all $x$. Using the ratio test I get:
$\lim_{n \to \infty} \left|\dfrac{f_{n+1}(x)}{f_n(x)}\right| = \lim_{n \to \infty} \left| \dfrac{(-1)^{n+1} \, x^{2n+3}}{2^{n+1} \, (n+1)!}\; \dfrac{2^n \, n!}{(-1)^n \, x^{2n+1}}\right|$
This simplifies to:
$\lim_{n \to \infty} \left|\dfrac{x^2}{2 \, (n+1)}\right|$
Now if $x \to \infty$, this limit won't be less than 1, right?
However, on the internet I found some pages saying the considered sum absolutely converges for all $x$ $\implies$ radius of convergence is $\infty$.
Did I do something wrong?