Does $\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$ converge absolutely?

Question

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?

Answer 1

No, it doesn't depend on $x$. It always converges absolutely. Note that, if $x\neq0$,$$\frac{\frac{|x|^{2k+2}}{(2k+5)!}}{\frac{|x|^{2k}}{(2k+3)!}}=\frac{|x|^2}{(2k+5)(2k+4)}\to0.$$

Answer 2

$\sum_{k=2}^{\infty} \dfrac {x^{2k}}{(2k+3)!} \lt$

$\sum_{k=2}^{\infty}\dfrac{x^{2k}}{(2k)!} \lt$

$\sum_{k=0}^{\infty}\dfrac{|x|^k}{k!} =e^{|x|}.$

For $x \in \mathbb{R}$ the absolute partial sums are mon. increasing and bounded above by $e^{|x|},$

hence convergent for $x \in \mathbb{R}.$

Answer 3

Yes since $$\sinh x = \sum\limits_{k= 0}^\infty\frac{x^{2k+1}}{(2k+1)!}$$

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