Why $1-\frac{1}{2!}+\frac{1}{3!}-...+(-1)^{n+1}\frac{1}{n!}\to 1-\frac{1}{e} $

Question

$$1-\frac{1}{2!}+\frac{1}{3!}-...+(-1)^{n+1}\frac{1}{n!}$$ $$\to 1-\frac{1}{e} $$ as $$n \to \infty$$

i know $\sum_{n=1}^{\infty} \frac{1}{n!}=e$, but this is alternating

Answer 1

The Taylor series for $e^x$ is $\sum_{i=0}^\infty \frac {x^i}{i!}$ If we evaluate this at $x=-1$ we get $\frac 1e=\sum_{i=0}^\infty \frac {(-1)^i}{i!}$ From this we get $$1-\frac 1e=1-\sum_{i=0}^\infty \frac {(-1)^i}{i!}\\ =\sum_{i=1}^\infty\frac {(-1)^{i-1}}{i!}$$ by canceling the $i=0$ term and negating the rest

Answer 2

Note that by changing $x$ to $-x$ we have $$ e^{-x} = 1 -x + (-x)^2 /{2!} + (-x)^3 / {3!} +.....$$

Let $x=1$ and you have the result.