We need to find the value of the infinite sum $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$.

1. Problem Description

We need to find the value of the infinite sum

$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$.

2. Solution Steps

We know the Taylor series expansion for $e^x$ is given by:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

Now, let's consider the sum $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n!}$.

We can rewrite $(-1)^{n+1}$ as $-(-1)^n$. Then the sum becomes:

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

The Taylor series for $e^x$ starts at $n=0$. In our case, the sum starts at $n=1$. Thus,

$e^x = \frac{x^0}{0!} + \sum_{n=1}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$

So, $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$

Substituting $x = -2$, we have:

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

Therefore,

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

3. Final Answer

$1 - \frac{1}{e^2}$