Identification of power series

Question

Using known power series, I am asked to identify the following function:

$y= \sum_{n=1}^\infty \frac{(-1)^n2^n}{n!} x^n$

So far, knowing that $e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1+x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$, I can rewrite my function to be $y= e^2\sum_{n=1}^\infty (-1)^n -1$

Is there a known function for the sum of (-1)^n, or does my solution seem sufficient?

Answer 1

$$\sum_{n=1}^\infty\frac{(-1)^n2^n}{n!}x^n = \sum_{n=0}^\infty\frac{(-2x)^n}{n!} -1 = e^{-2x}-1$$

Answer 2

You identified the power series right! The power sum to your problem should be $$e^{-2x} - 1$$