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So I have to prove 2 things:

  1. That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R, x>0$.

  2. That $\lim\limits_{n \rightarrow \infty}\frac{x^n}{n!} = 0$ where $n \in \mathbb N$ and $x \in \mathbb R$.

For #1, I know that $\frac{x^n}{n!} >0$, which means that I can find an upper bound and use squeeze theorem. For #2, I have no idea where to start.

BigSpicyPotato
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  • For (2), use the fact that $|a_{n}|\rightarrow 0$ implies $a_{n}\rightarrow 0.$ – RideTheWavelet Sep 27 '17 at 17:03
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    $c_n = \frac{x^n}{n!}$, $|c_{n+k}| \le |c_n| , |x/n|^k$ – reuns Sep 27 '17 at 17:06
  • Depending on what you have available for you to use, you can prove it via the convergence of the MacLaurin series for $e^x$ for any $x\in\Bbb R$ (where this convergence can be simply proven by the ratio test). See my answer on this question: https://math.stackexchange.com/questions/2129489/limit-nx-x-as-x-tends-to-infty/2129500#2129500 – Dave Sep 27 '17 at 17:13

6 Answers6

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I think I've answered this question before, but note that $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$, so for every fixed $x\in \mathbb{R}$ the series converges, thus the sequence of the terms $\{\frac{x^n}{n!}\}_{n\in \mathbb{N}}$ must converge to zero as $n$ goes to infinity, otherwise, the sum would not converge.

Koto
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  • It is the opposite : you prove $f(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$ converges and $f'(x) = f(x)$ so that $f$ is an exponential to obtain $e^x = f(x)$ – reuns Sep 27 '17 at 17:08
  • @reuns, not really, I used the fact that $f(x)$ converges to prove that the sequence of terms cannot go to anything but zero. Suppose that the limit is not zero, then $f(x)$ cannot be convergent. – Koto Sep 27 '17 at 17:10
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    This logic is fine, but you may as well include a proof that the series converges, to show that this isn't a circular argument. –  Sep 27 '17 at 17:45
  • @Bungo, I think this is nice: https://math.stackexchange.com/questions/2136683/convergence-of-the-ex-taylor-series – Koto Sep 27 '17 at 17:47
  • Well, you don't need the fact that the series equals $e^x$. It suffices to show that the series converges for all $x$. The ratio test makes this very easy. –  Sep 27 '17 at 17:49
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Hint: Given $x$, take $N \in \mathbb{N}$ such that $N>x$. Then, for $n >N$, $$\frac{x^n}{n!} = \frac{x^N}{N!} \cdot \frac{x^{n-N}}{(n!/N!)} \le \frac{x^N}{N!} \cdot \left(\frac{x}{N+1}\right)^{n-N} .$$

Zestylemonzi
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A little bit of support for Koto:

For every $x \in \mathbb{R}$ the exponential series

$\star)$ $\exp(x): = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ converges absolutely.

Hence: $\lim_{n \rightarrow \infty} |\dfrac{x^n}{n!}| = 0$ for $x \in \mathbb{R}$.

Proof of $\star$):

Ratio test:

With $a_n: = \dfrac{x^n}{n!}$ we get for

$x\ne 0$ and $n\ge 2|x|$:

$|\dfrac{a_{n+1}}{a_n}| =$

$ |\dfrac{x^{n+1}}{(n+1)!}\dfrac{n!}{x^n}|=$

$\dfrac{|x|}{n+1}\le 1/2$.

Peter Szilas
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First Answer The series $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ converges then $$\frac{x^n}{n!}\to 0$$

OR Second Answer Use the following famous Stirling formula: Given $x>0$ $$ \lim_{n\to +\infty} \frac{n!}{\left(\frac{n}{e}\right)^n\sqrt{2n} }=\sqrt{\pi}. $$ and $$|x^n| =e^{n\ln |x|}$$

Guy Fsone
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  • Stirling's Formula is a bit of overkill here. Note that $n!\ge (n/2)^{n/2}$, which can be shown using elementary analysis. – Mark Viola Sep 27 '17 at 18:20
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Note that $n!\ge (n/2)^{n/2}$. Then, we have

$$\left|\frac{x^n}{n!}\right|\le \frac{|x|^n}{\left(\sqrt{n/2}\right)^n}=\left(\frac{2|x|}{n}\right)^n$$

Can you conclude now?

Mark Viola
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  • @markdodds Mark, is this useful? Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Sep 28 '17 at 00:17
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$$ \frac {x^n}{n!} = \frac{\overbrace{x\cdot x\cdot x\cdots\cdots x}^{\Large n \text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots\cdots n}_{\Large n \text{ factors}}} $$ When $n$ gets to be more than twice as big as $x,$ then multiplying by $\dfrac x n$ reduces the thing to less than half what it was. Then $n$ keeps growing (while $x$ does not) and at some point $n$ is more than three times as big as $x$ and then you're multiplying by $\dfrac x n < \dfrac 1 3$ and making the thing at each step less than $1/3$ what it had been. And so on$\,\ldots$ So it approaches $0.$

(That assumes $x$ is positive. If $x$ is negative, then think about $\dfrac{|x|^n}{n!}.$ It that approaches $0,$ then without the absolute value that still approaches $0.$

stoic-santiago
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Michael Hardy
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