Exponential Series

$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots$ equals

If $y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$, then $x=$

If $s=\sum _{n=0}^{\infty }\frac{{\left({\log }_{e}x\right)}^{2n}}{\left(2n\right)!},$ then $s=$

Sum to infinity of the series is $1+\frac{x^2}{2!}+\frac{x^4}{4!}+ \dots$ is

$1+\frac{a-bx}{1!}+\frac{{\left(a-bx\right)}^2}{2!}+\frac{{\left(a-bx\right)}^3}{3!}+\dots \infty =$

$3+\frac{5}{1!}+\frac{7}{2!}+\frac{9}{3!}+\dots \infty =$

$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\dots \infty =$

The value of ${\left[1+\frac{1}{2!}+\frac{1}{4!}+\dots \right]}^2-{\left[1+\frac{1}{3!}+\frac{1}{5!}+\dots \right]}^2$ is 

If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \infty ,$ then $x=$