Evaluation of series $\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$

How to evaluate series $$\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$$ I tried to split the summation...but I failed. Please help

Hint: $$5n+1 = \frac{5}{2}(2n+1) - \frac{3}{2}$$
Do you know $$\sum_{n=0}^\infty \frac{1}{(2n+1)!}$$ and $$\sum_{n=0}^\infty \frac{1}{(2n)!}$$ ?
• Thank you. We can group the terms accordingly to get the answer $\frac{e}{2}+\frac{2}{e}$ – user538954 May 21 at 16:44
$$\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$ $$\frac{1}{x}\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n+1)!}$$ let $$x\rightarrow \sqrt{x}$$ $$\frac{1}{\sqrt{x}}\sinh \sqrt{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{(2n+1)!}$$
let $$x\rightarrow x^5$$ $$\frac{1}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n}}{(2n+1)!}$$ multiply by $$x$$ $$\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n+1}}{(2n+1)!}$$ $$(\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5})'=\sum_{n=0}^{\infty}\frac{(5n+1)x^{5n}}{(2n+1)!}$$ then let $$x=1$$ to get the sum