Prove that $\lim_{s \to \infty} \sum_{x=1}^{2s} (-1)^x\sum_{n=1}^{x}\frac{1}{n!}=\cosh (1) -1$ How can we prove that $$\lim_{s \to \infty} \sum_{x=1}^{2s} (-1)^x\sum_{n=1}^{x}\frac{1}{n!}=\cosh (1) -1$$

It seems like this is some kind of telescopic series, but I don't know how to find the limit of this sum. Any help would be greatly appreciated.
HINT:

Note that we can write 

$$\begin{align}
\sum_{x=1}^{2s}(-1)^x \sum_{n=1}^x\frac1{n!}&amp;=\sum_{x=1}^s\left(-\sum_{n=1}^{2x-1}\frac1{n!}+\sum_{n=1}^{2x}\frac1{n!}\right)\\\\
&amp;=\sum_{x=1}^s\frac1{(2x)!}
\end{align}$$
$$\cosh(x)=\frac{e^x+e^{-x}}{2}\Rightarrow \cosh(1)=\frac{e+e^{-1}}{2}$$



$$S=\sum_{x=1}^{2s}(-1)^x\sum_{n=1}^x\frac{1}{n!}$$
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\Rightarrow \cosh(1)=\frac 12\sum_{n=0}^\infty\frac{1^n+(-1)^n}{n!}=\frac{1}{2}\sum_{n=0}^\infty\frac{\left[1+(-1)^n\right]1^n}{n!}$$
now since for even odd n the terms will equal zero, we want:
$$\cosh(1)=\sum_{n=0}^\infty\frac{1^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{1}{(2n)!}$$
going back to $S$ now we have:
$$S=\sum_{x=1}^{2s}(-1)^x\sum_{n=1}^x\frac{1}{n!}=-\left(\frac 1{1!}\right)+\left(\frac 1{1!}+\frac 1{2!}\right)-\left(\frac 1{1!}+\frac 1{2!}+\frac 1{3!}\right)+...$$
and if you look at which terms you will cancel you notice that we are left with:
$$S=\sum_{x=1}^s\frac 1{(2x)!}$$
Now we are simply left with:
$$\lim_{s\to\infty}S_s=\cosh(1)$$