The value of $$\sum_{k=1}^{\infty}(-1)^{k+1}\left(\frac{k(k+1)}{k!}\right)$$ is

We need to evaluate $$\sum_{k=1}^{\infty}(-1)^{k+1}\frac{k(k+1)}{k!}$$.

$$\frac{k(k+1)}{k!} = \frac{k+1}{(k-1)!} = \frac{k-1+2}{(k-1)!} = \frac{1}{(k-2)!} + \frac{2}{(k-1)!}$$ (for $$k \geq 2$$)

For k=1: $$(-1)^2 \cdot \frac{1 \cdot 2}{1!} = 2$$

For $$k \geq 2$$:

$$\sum_{k=2}^{\infty}(-1)^{k+1}\left(\frac{1}{(k-2)!} + \frac{2}{(k-1)!}\right)$$

Let m = k-2 for the first part: $$\sum_{m=0}^{\infty}\frac{(-1)^{m+3}}{m!} = -\sum_{m=0}^{\infty}\frac{(-1)^m}{m!} = -e^{-1}$$

Let m = k-1 for the second part: $$2\sum_{m=1}^{\infty}\frac{(-1)^{m+2}}{m!} = 2\sum_{m=1}^{\infty}\frac{(-1)^m}{m!} = 2(e^{-1} - 1)$$

$$S = 2 + (-e^{-1}) + 2(e^{-1} - 1) = 2 - e^{-1} + 2e^{-1} - 2 = e^{-1} = \frac{1}{e}$$

Therefore, the answer is Option 3: $$\frac{1}{e}$$.