# Show that $\lim_{n\rightarrow\infty}e^{-n}=0$

I wrote out the term from above and get

$$\text{Show that: }\lim_{n\rightarrow\infty}\sum_{k=0}^{\infty}(-1)^k\frac{n^k}{k!}=0$$

I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $$0$$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?

Edit: I have changed the fraction in the expression

• $e^{-n} = \frac{1}{e^n}$ also $2<e$ and $2^n\rightarrow\infty$ so $\frac{1}{2^n}\rightarrow 0$ and $1/e<1/2$ – Yanko Jan 23 at 14:27
• This is just a special case of the general fact: if $|x|<1$ then $\lim_{n \to \infty} x^n=0$. – Lee Mosher Jan 23 at 14:39
• Although $\lim_{n\rightarrow\infty}\sum_{k=0}^{\infty}(-1)^k\frac{n^k}{k!}=0$ is true, you cannot prove it by looking at the terms of this series. – GEdgar Jan 23 at 15:04

Here is the standard elementary proof that $$a^n \to 0$$ if $$0 < a < 1$$.

$$a =1/(1+b)$$ where $$b = 1/a -1 > 0$$.

By Bernoulli's inequality, $$(1+b)^n \ge 1+bn > bn$$ so $$a^n =1/(1+b)^n < 1/(bn) \to 0$$ as $$n \to \infty$$.

For any $$a$$ such that $$0 < a < 1$$, you can find natural $$n$$ to make $$a^n$$ smaller than any positive number $$x$$:

$$a^n < x \iff n \, log(a) < log(x) \iff n > log(x)/log(a)$$

Also, $$a^n > 0$$ for any natural $$n$$.

These 2 facts imply $$lim_{n -> \infty} \, a^n = 0$$

Do not expand, just use

$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.