Find $f(0)$ if $f(x)= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!} $ 
Find $f(0)$ if $f(x)= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!} $

This seems trivial but I am not able to figure it out. Do I directly put $x=0$ or should I reindex $n=0$ and use Taylor Theorem. If I use Taylor theorem, then is $f(0)$ simply the $n$ part of the numerator?

Hint
$$f(x)= \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n-2}}{(2n+1)!}=\frac{1}{3!}-\frac{x^2}{5!}+\cdots$$
