How does this pattern work?

Question

I know that $$ \sum_{k=0}^{\infty}\frac{1}{k\,!}=e=\lim_{n \to \infty}{(1+\frac{1}{n})^n} $$ but why $$ \sum_{k=0}^{\infty}\frac{1}{(2k)\,!} = \cosh(1) $$ and $$ \sum_{k=1}^{\infty}\frac{1}{(2k+1)\,!} = \sinh(1)-1 $$ and $$ \sum_{k=2}^{\infty}\frac{1}{k\,!} = e-2 $$ It seems that we can get various results by changing the starting value of k and the denominators.
My question is that is there a formula or a theorem to explain this pattern?

Answer 1

We have

$$\cosh(1)=\frac{e^1+e^{-1}}{2}=\frac12\sum_{k=0}^\infty\frac{1+(-1)^k}{k!}=\frac12\sum_{k=2p=0}^\infty\frac{2}{(2p)!}=\sum_{p=0}^\infty\frac1{(2p)!}$$

and by the same method we get the expression of $\sinh(1)$.