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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

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    $\begingroup$ $e^{-n} = \frac{1}{e^n}$ also $2<e$ and $2^n\rightarrow\infty$ so $\frac{1}{2^n}\rightarrow 0$ and $1/e<1/2$ $\endgroup$ – Yanko Jan 23 at 14:27
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    $\begingroup$ This is just a special case of the general fact: if $|x|<1$ then $\lim_{n \to \infty} x^n=0$. $\endgroup$ – Lee Mosher Jan 23 at 14:39
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    $\begingroup$ 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. $\endgroup$ – GEdgar Jan 23 at 15:04
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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 $n \to \infty$.

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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 $lim_{n -> \infty} \, a^n = 0$

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Do not expand, just use

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

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