Finding $\lim_{x\to\infty} \frac {(x!)^{\frac 1 x}}{x}$

Question

Find $\displaystyle \lim_{x\to\infty} \frac {(x!)^{\frac 1 x}}{x}$

I have no idea how to solve it, I can approximate it to be in $(0,1)$ by squeezing but getting to the solution $(\frac 1 e)$ seems like it would require a lot more. Is this an identity?

Note: no integrals nor gamma function.

I feel like this one has come up a couple of times recently. Anybody find a dup? — Thomas Andrews, Apr 12 '15 at 17:38
Yeah, I'm not blaming you. This is the sort of problem that is hard to find duplicates of, because it is heavily notation-dependent. Change $x$ to $y$, for example, and the search algorithm doesn't match the question. — Thomas Andrews, Apr 12 '15 at 17:41
Maybe you can search for Stirling's formula. It will probably be included somewhere in the duplicate. — mickep, Apr 12 '15 at 17:41
So does it have to be using approximation tools like Stirling's or Riemann sums? — shinzou, Apr 12 '15 at 17:47
Possible duplicates of this question, or this question, or this question. There may be others. — robjohn, Apr 12 '15 at 20:59

score 4 · Answer 1 · edited Apr 12 '15 at 18:23

4

While waiting for someone to find a duplicate a reminder of Stirling formula

$$x\to\infty\implies x!\sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$$

So we have, when $x\to\infty$,

$$\frac{(x!)^{\frac{1}{x}}}{x}\sim \frac{1}{e}\left(\sqrt{2\pi}\right)^{\frac{1}{x}}x^{\frac{1}{2x}}\sim \frac{1}{e}$$

edited Apr 12 '15 at 18:23

Prasun Biswas

6,334

answered Apr 12 '15 at 17:59

marwalix

16,773

science · Accepted Answer · 2015-04-12T18:48:18.977

2

Note this $$ \left( \frac{x!}{x^x} \right)^{1/x} = (a_x)^{1/x} $$

where $a_x = \frac{x!}{x^x}$ and then use the fact that

$$ \lim_{x\to \infty} (a_x)^{1/x} = \lim_{x\to \infty} \frac{a_{x+1}}{a_x} $$

and the evaluation of limit will become easy

$$ \lim_{x\to \infty} \frac{a_{x+1}}{a_x} = \lim_{x\to \infty} \frac{1}{(1+1/x)^x} = \frac{1}{e}. $$

edited Apr 12 '15 at 18:48

answered Apr 12 '15 at 17:52

science

2,900

But what do you do with rest? i.e: $\frac {((x+1)!)^{1/x+1}}{(x!)^{1/x}(x+1)}$? – shinzou Apr 12 '15 at 17:59
Ah you took out the power. Nice. – shinzou Apr 12 '15 at 18:00
@kuhaku: Yes in order to match a certain form. – science Apr 12 '15 at 18:01
I never saw it being done, is it allowed because both sides are equal to 1? – shinzou Apr 12 '15 at 18:04
Is it a general theorem that $\lim a_n = \lim (a_n)^{1/n} = \lim \frac {a_{n+1}} {a_n}$? But from limit arithmetic both of the last terms should be $1$... – shinzou Apr 12 '15 at 18:11
@kuhaku: I used the result $ \lim (a_n)^{1/n} = \lim \frac {a_{n+1}} {a_n} $ which is a well known result and your $a_n= \frac{n!}{n^n}$. I think this is clear. So no integrals no gamma function in this approach. – science Apr 12 '15 at 18:34
But why is it equal to $\lim a_n$? – shinzou Apr 12 '15 at 18:43
@kuhaku: your limit problem is $ \lim_{n\to \infty} (a_n)^{1/n} $ where $a_n = \frac{n!}{n^n}$. – science Apr 12 '15 at 18:46

score 2 · Answer 3 · 2015-04-12T20:21:16.333

Just for variety, here's an alternative solution, based on the first simple observation in Terry Tao's notes on Stirling's formula.

We'll show that $$ \frac{n^n}{n!} \le e^n \le 2(n+1)\frac{n^n}{n!} \tag{$\ast$} $$ Taking $n$th roots and squeezing yields the desired limit. (The standard Riemann sum method — as, e.g., in the first stage in Tim Gowers' "Removing the magic from Stirling's formula" — gives the similar bounds $e\frac{n^n}{n!} \le e^n \le en\frac{n^n}{n!}$.)

Tao's observation gives us the lower bound in ($\ast$): $$ \frac{n^n}{n!} \le \sum_{k=0}^\infty \frac{n^k}{k!} = e^n $$ Now let's consider how much we lost there by throwing away all terms of the series except the one with $k=n$. First, $$ \frac{n^k}{k!} \le \frac{n^{k+1}}{(k+1)!} \iff k\le n $$ and so the terms in the series $\sum_{k=0}^\infty \frac{n^k}{k!}$ increase from $k=0$ to $k=n$ and decrease thereafter. In particular, for a longish head of the series we get $$ \sum_{k=0}^{2n-1} \frac{n^k}{k!} \le 2n\cdot\frac{n^n}{n!} \tag{1} $$ In the remaining tail, we have $k\ge 2n$, and so $$ \frac{n^{k+1}}{(k+1)!} = \frac{n}{k+1}\cdot\frac{n^k}{k!} \le \frac12\cdot\frac{n^k}{k!} $$ so each term in the tail is less than half the preceding one. Thus we can bound them by the geometric series $1+\frac12+\frac14+\dotsb$, getting $$ \sum_{k=2n}^\infty \frac{n^k}{k!} \le 2\frac{n^{2n}}{(2n)!} \le 2\frac{n^n}{n!} $$ which when combined with (1) gives the upper bound in ($\ast$).

score 2 · Answer 4 · answered Apr 12 '15 at 21:17

Note that $$ \frac{(n+1)!}{n!}=n+1 $$ and $$ \frac{(n+1)^{n+1}}{n^n}=\left(1+\frac1n\right)^{n}(n+1) $$ Therefore, $$ \left.\frac{(n+1)!}{(n+1)^{n+1}}\middle/\frac{n!}{n^n}\right.=\left(1+\frac1n\right)^{-n} $$ Since $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^{-n}=e^{-1}$, Stolz-Cesaro applied to the logarithms says that $$ \lim_{n\to\infty}\frac{(n!)^{1/n}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{1/n}=e^{-1} $$

Prasun Biswas · Answer 5 · 2015-04-12T19:00:44.443

1

Let us first consider the limit as $L$. Then, applying natural logarithm, we get,

$$\ln(L)=\lim_{x\to\infty}\frac{1}{x}\sum_{i=1}^x\left(\ln\left(\frac{i}{x}\right)\right)=\int\limits_0^1\ln(x)\,dx$$

Now, since the function $f(x)=\ln(x)$ is Riemann-integrable on $(0,1)$, we see, by the Riemann Sum argument that $L$ exists finitely. Next, we start to evaluate it.

$$L=\int\limits_0^1\ln (x)\,\mathrm dx=\left[x\ln(x)-x\right]_0^1\\ \implies \ln(L)=\left(\ln(1)-1\right)-\left(\lim_{x\to 0^+}(x\ln(x)-x)\right)=(-1)\\ \implies \ln(L)=(-1)\implies L=e^{-1}=\frac{1}{e}$$

edited Apr 12 '15 at 19:00

answered Apr 12 '15 at 17:51

Prasun Biswas

6,334

1

This is nice but should be stated in another way. By saying "take the limit as $L$" we are assuming that the limit exists. It exists quite trivially (since $\log x$ is a Riemann-integrable function over $(0,1)$), but a better choice for wording would have been "since the following Riemann sum argument shows that [...] we have that the limit equals $\frac{1}{e}$". – Jack D'Aurizio Apr 12 '15 at 18:43
1

@JackD'Aurizio, I have edited my answer. Can you please verify it for me? I'm actually still learning Calculus, so I'm not familiar with many terms, like the term "Riemann integrable", etc. If you find any more errors, it'd be kind of you if you could notify me. Thanks. :) – Prasun Biswas Apr 12 '15 at 19:02

Finding $\lim_{x\to\infty} \frac {(x!)^{\frac 1 x}}{x}$

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