Why is $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$? In order to sum up the series we have to realize that it is almost the expansion of the cosine function. But we need $k=0$ in the sum. I don't know how to do this. The solution is   $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=\left.\left(\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k) !} x^{2 k}-1\right)\right|_{x=1}$, but where does $-1$ on the right hand side come from and how did we achieve $k=0$?
My approach was to do the following, which leads to a dead end...(right?)
$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{(2 k) !}=-\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2 k) !}=\cdots$

As @Chaos & @Fred said, it's because of first term added to right hand side ($k=0$). Also notice the benefit of relation:
$$f(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k} \Longrightarrow
f''(x) = -f(x) \Longrightarrow f(x) = a \sin (x) + b \cos (x)$$
May this is the purpose of relation.
