Does $\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$ converge absolutely? Let $\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$, $x\in\mathbb{R}$. Does this series converge absolutely?
$$\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$$ converges absolutely if $$\sum\limits_{k=2}^\infty\left|\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}\right|=\sum\limits_{k=2}^\infty\frac{|x|^{2k}}{(2k+3)!}$$ converges.
I guess that it depends on $x$ if this converges or not. For $x\in [-1,1]$ by the ratio test.
However, I'm not sure about the other $x$. How to proceed further?

Yes since $$\sinh x = \sum\limits_{k=
0}^\infty\frac{x^{2k+1}}{(2k+1)!}$$
then $$  \sum\limits_{k=
2}^\infty\frac{x^{2k}}{(2k+3)!}\le  \sum\limits_{k=
2}^\infty\frac{x^{2k}}{(2k+1)!}= -1-\frac{x^2}{6}+ \sum\limits_{k=
0}^\infty\frac{x^{2k}}{(2k+1)!}\\=-1- \frac{x^2}{6}+ \frac{1}{|x|}\sum\limits_{k=
0}^\infty\frac{|x|^{2k+1}}{(2k+1)!}= =-1- \frac{x^2}{6}+ \frac{|\sinh x|}{|x|} $$
