What is the limit of $(x!)^{1/x}$ as $x$ approaches infinity?

Question

I'm doing Sequences and Series at the moment and I came upon this limit:

$$\lim_{x \to ∞} (x!)^{1/x}$$

Here is how I evaluated it:

$$y = \lim_{x \to ∞} (x!)^{1/x}$$ $$\ln{y} = \lim_{x \to ∞} \ln({(x!)^{1/x}})$$ $$\ln{y} = \lim_{x \to ∞} \frac{\ln(x!)}{x}$$ $$\ln{y} = \lim_{x \to ∞} \frac{lnx + \ln(x-1) + \ln(x-2) +...+ \ln(2) + \ln(1)}{x}$$

we have the indeterminate form $\frac{∞}{∞}$, so we can use L'Hopital's Rule:

$$\ln{y} = \lim_{x \to ∞} \frac{\frac{1}{x} + \frac{1}{x-1} + \frac{1}{x-2} +...+ 0 + 0}{1}$$ $$\ln{y} = 0$$ $$y=1$$

However, this seems to be wrong just from numerical analysis and from what I've found online. What is my error? Thank you very much for any help!

Answer 1

The "What's wrong" part is when you take derivative of $\ln(x)+\ln(x-1)+\cdots +\ln(2)+\ln(1)$.

Consider taking derivative of $x=1+1+1+\cdots+1+1$ with $x$ terms. You will get $1=0+0+0+\cdots+0+0$, which is of course wrong. This is because the right hand side is dependent on $x$.

To solve your limit, I'll continue from $\lim_{x \to \infty}\frac{\ln(x!)}{x}$.

$$\lim_{x\to\infty}\frac{\ln(x!)}{x} \sim \lim_{x\to\infty}\frac{x\ln(x)-x}{x}=\lim_{x\to\infty}(\ln(x)-1)=+\infty$$

Answer 2

Your error lies in your use of L'Hopital's Rule. That would have worked if the numerator was a sum with a fixed number of terms, but it is not. By the same argument, we would have\begin{align*}1&=\lim_{n\to\infty}\frac nn\\&=\lim_{n\to\infty}\frac{1+1+1+1+\cdots+1}n\\&=\lim_{n\to\infty}\frac{0+0+0+0+\cdots+0}1\\&=0.\end{align*}

You will find a correct solution here.

Answer 3

Another approach. When $x$ is even, the product $x!$ has $x/2$ factors $ \gt x/2$, and the other factors are $\ge 1$. So $$ x! \ge \left(\frac{x}{2}\right)^{x/2} \\ (x!)^{1/x} \ge \left[\left(\frac{x}{2}\right)^{x/2}\right]^{1/x} =\left[\left(\frac{x}{2}\right)^{1/2}\right] = \frac{\sqrt{x}}{\sqrt{2}} . $$ But $\sqrt{x}/\sqrt{2} \to \infty$ as $x \to \infty$, so $(x!)^{1/x} \to \infty$.

The case of odd $x$ is similar.

Answer 4

What is my error?

Your error is twofold, depending on how you're defining $x!$. The original question makes sense if you take the limit across integers $x$ as $x \to \infty$, and we know what $x!$ is. But to take a derivative of $\ln x!$, you need $x!$ to be continuous (and differentiable) --- and $x!$ is not obviously differentiable.

Thus your first error is in thinking that you can differentiate $\ln x!$ as you have done.

But in fact it is possible to write a continuous analog of $x!$ (namely $\Gamma(x + 1)$, which is differentiable. The derivative isn't the same as what you wrote, but this is a path forward.

Your second error (assuming your earlier differentiation step had worked out) is in thinking that $$ \lim_{x \to \infty} \frac{1}{x} + \frac{1}{x - 1} + \cdots + \frac{1}{1}$$ tends to zero. In fact you have $x$ terms at least as large as $1/x$, and thus the limit is at least $1$ (but really is much larger than that). But even this limit should make you pause --- for each value of $x$, there are a different number of terms in the limit. That's suspicious and confusing --- and this is likely where the intuition gap lies.

Answer 5

Whenever I see a problem of limits where factorials are present, my first (and generally last) reaction is th think about Stirling approximation (just as @Gareth Ma commented).

As you wrote it $$y=(x!)^{\frac{1}{x}}\implies \log(y)=\frac{1}{x}\log(x!)$$ Now, using Stirling approximation $$\log(x!)=x (\log (x)-1)+\frac{1}{2} \log (2 \pi x)+O\left(\frac{1}{x}\right)$$ $$\log(y)=\log (x)-1+\frac{1}{2x} \log (2 \pi x)+O\left(\frac{1}{x^2}\right)$$ $$y=e^{\log(y)}=\frac{x}{e}+\frac{\log (2 \pi x )}{2 e}+O\left(\frac{1}{x}\right)$$ Use your pocket calculator with $x=1000$; it would give $y=369.492$ while the truncated expression written above gives $369.488$.