Calculate $\lim_{n \to \infty}\frac{a^n}{n!}$

Question

Calculate $$\lim_{n \to \infty}\frac{a^n}{n!}$$

Attempt

Consider $$\lim_{n \to \infty}\exp(\log(\frac{a^n}{n!}))=\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)$$

for $a>1$ $$ \lim_{n \to \infty}\left(n\log(a)\right)=\infty $$ $$ \sum_{n \geq 1}\log(n)\leq \int_{1}^{\infty}{\log(x)dx}=\infty$$

therefore $$\exp\left(\lim_{n \to \infty}\left(n\log(a)-\sum_{n\geq 1}\log(n)\right)\right)=\exp(-\infty)=0$$ therefore

$$\lim_{n \to \infty}\frac{a^n}{n!}=0$$ Is my proof right?

I think that is not general, since $a>1$ and in the problem $a$ is arbitrary

Are you assuming from the outset that the limit actually exists? — Umberto P., Feb 26 '21 at 20:37
Your sum index is $n$. Not good. Choose $k$ and let the upper bound for the sum be $n$. — vitamin d, Feb 26 '21 at 20:38
Not, because the limit exists iff the other series of limits exists — Juan T, Feb 26 '21 at 20:38
The line after consider is not valid. You defined $n$ as a limit variable but now it is also a sum index?? — vitamin d, Feb 26 '21 at 20:40
HINT: $$\left|{a^n\over n!}\right|\le k {|a|\over n}$$ for large enough $n$, where $k$ is a constant. — Intelligenti pauca, Feb 26 '21 at 20:44

score 3 · Accepted Answer · answered Feb 26 '21 at 20:50

3

Replying to your last comment: Yes there is a faster way. Show that for sufficiently large $n$ the following inequality holds: $$n!>(a+1)^n.$$ We now have $$\lim_{n\to\infty} \frac{a^n}{n!} \leq \lim_{n\to\infty} \frac{a^n}{(a+1)^n} = \lim_{n\to\infty} \left(\frac{a}{a+1}\right)^n \to 0.$$

answered Feb 26 '21 at 20:50

vitamin d

5,783

It´s a nice proof – Juan T Feb 26 '21 at 20:51
it´s relation with binomial expression since $a^n=(1+(a-1))^n$? – Juan T Feb 26 '21 at 20:52
I don't understand. What is your question? Your equality is right but why is it important? – vitamin d Feb 26 '21 at 20:53
exists a relation between $n!> (a+1)^n$ and the expression above? – Juan T Feb 26 '21 at 20:54
Ohh. There is a nice way to think about it. I will show it for $a=1$. You have to prove it then for all $a\in\mathbb{R}$. $$1\cdot2\cdot3\cdot4\cdots(n-1)\cdot n > \underbrace{2\cdot2\cdot2\cdot2\cdots2\cdot2}_{\text{$n$ times}}$$ – vitamin d Feb 26 '21 at 20:57
Thanks a lot!! for the idea – Juan T Feb 26 '21 at 20:58

score 1 · Answer 2 · answered Feb 26 '21 at 21:05

Here is an alternative:

It seems you are working with $a>1$, although it is not stated at the beginning but rather in the middle. I will go with that assumption:

$a^n>0$ and $n! > 0, \forall n \in \mathbb{N}$.

Now,

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n} $

At some point we will have $n > a$, say this occurs at $N$, take $n > N$ then

$\displaystyle \frac{a^n}{n!} = \frac{a\cdot a \cdots a}{1\cdot 2 \cdots n} \le \frac{a \cdots a}{1 \cdots (N-1)}\cdot \frac{a \cdots a}{N \cdots n} \le \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n}$

Now, since we have $m > a, \forall m \in \{N,\ldots,n\}$ we know $\displaystyle \frac{a}{m} < 1$

Hence

$\displaystyle \frac{a^{N-1}}{(N-1)!}\cdot \frac{a^{n-(N-1)}}{N\cdots n} \le \frac{a^{N-1}}{(N-1)!}\cdot \frac{a}{n} \to 0$ as $n \to \infty$

score 0 · Answer 3 · answered Feb 26 '21 at 20:43

0

Hint : Starting from your idea, try to find the limit of $n log(a) - \sum_{k=1}^n log(k)$. Then apply what you know about the $exp$ function.

answered Feb 26 '21 at 20:43

FiMePr

1,219

The exponential function separe the product – Juan T Feb 26 '21 at 20:46
And tends to 0 at $-\infty$. – FiMePr Feb 28 '21 at 10:19

Calculate $\lim_{n \to \infty}\frac{a^n}{n!}$

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