The sum of the series $1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\dots$ upto infinity is:

The expansion of $e^x$ is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$.For $e^{-1}$, substitute $x=-1$:
$$e^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dots$$
$$e^{-1}=0+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dots$$
The required series is $S=1-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\dots$.We can see that $S=1-(\frac{1}{2!}-\frac{1}{3!}+\dots )=1-e^{-1}$.Wait, re-checking the logic in the prompt's snippet:The snippet says $1-\frac{1}{1!}+\frac{1}{2!}\cdots =e^{-1}$ (which is 0).Looking at the specific series: $1-\frac{1}{2!}+\frac{1}{3!}\dots$. This is $1-(1-e^{-1})=e^{-1}$. (Since $e^{-1}=1-1+1/2!-1/3!\dots$).So, $1-1/2!+1/3!-\cdots =1-(1-e^{-1})=e^{-1}$.