I want to prove that:

$$\sum\limits_{i=0}^\infty \frac{(-1)^i}{i!}=\lim_{i \to \infty}\bigg(1-\frac{1}{i}\bigg)^i$$

First I need to prove series is convergent. But the partial sums of the series is not monotone sequence.Can anyone tell me how to prove that it is convergent.I tried to show it is Cauchy sequence,but I couldn't. I'd appreciate any help.

## Best Answer

Use $(\rm i)$ Leibniz's Criterion, or $(\rm ii)$ a absolute convergence considerations.

$(\rm i)$ We have that $a_k=\dfrac 1 {k!}$ is monotone descreasing with $\lim\; a_k=0$, so Leibniz's criterion says $$\tag{1}\sum_{k=0}^\infty (-1)^k\frac 1 {k!}$$

converges.

$(\rm ii)$ $$\sum_{k=0}^\infty \frac 1 {k!}=e$$

so

$$\tag{1}\sum_{k=0}^\infty (-1)^k\frac 1 {k!}$$

converges, since it is absolutely convergent by $(1)$.