# Determine the sum $\sum_{n=0}^{\infty}{(-1)^n}\frac{2^{2n-1}}{(2n)!}$

I need to calculate the sum $\sum_{n=0}^{\infty}{(-1)^n}\frac{2^{2n-1}}{(2n)!}$ . It seems very "similar" to Taylor expansion of functions arcsin(x) and its derivative for x = -2.

It is known: $arcsin(x) = \sum_{n=0}^{\infty}\frac{{(2n-1)!!x^{2n+1}}}{2^{n}n!(2n+1)}$.

When derivative applied we get: $\frac{1}{\sqrt{1-x^{2}}} = \sum_{n=1}^{\infty}\frac{{(2n+1)(2n-1)!!x^{2n}}}{2^{n}n!(2n+1)} = \sum_{n=1}^{\infty}\frac{{(2n-1)!!x^{2n}}}{2^{n}n!}$.

What to do next? Am I on the wrong trace here?

• $\frac{cos(2)}{2}$ Jun 23 '15 at 1:14
• It's even more "similar" to, $$\cos(x)=\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$$ Divide it by $2$ and substitute $x=2$ to get your result (as shown by d.k.o) Jun 23 '15 at 1:41
It surprises me to see that you are familiar with the rather obscure series of $$\arcsin x$$, and yet, at the same time, fail to recognize one of the most well-known series in math, namely that of $$\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}~.~$$ Also, $$\sin x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}~.$$ Remember that $$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}~,~$$ and then recall Euler's formula $$e^{ix}=\cos x+i\sin x$$.