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I read that $\left(\frac{1}{n!}\right)^\frac{1}{n}$ goes to $0$ as $n$ goes to infinity.

I tried to solve this myself by expanding the factorial:

$$\left( \frac{1}{n!}\right)^\frac{1}{n}=\left( \frac{1}{n}\cdot \frac{1}{n-1}\cdot ... \cdot\frac{1}{1}\right)^\frac{1}{n}=\left( \frac{1}{n}\right)^\frac{1}{n}\cdot \left( \frac{1}{n-1}\right)^\frac{1}{n}\cdot ... \cdot\left( \frac{1}{1}\right)^\frac{1}{n}$$

The leftmost factor in the very right side converges to one, as does the rightmost, and the other factors should lie in between, hence also converge to one.

Then if I apply the limit rule stating that $\lim a\cdot b$ = $\lim a\cdot\lim b$ (with $a$ and $b$ being convergent sequences), I get that the whole thing converges to $1$, which I know it doesn't. My guess is that I cannot apply that limit rule for a product with an unbounded number of factors.

Can someone please give me a simple proof of convergence to $0$ and perhaps some insight on my mistake(s)? That would be lovely.

gt6989b
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Michele Mattei
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    The thing is that you cannot apply limit multiplication theorem infinity times. – Eugene Sirkiza Jan 08 '20 at 20:17
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    The problem with your rough heuristic is that you are assuming there are only finitely many terms, but the number of terms is also getting large! – A. Thomas Yerger Jan 08 '20 at 20:17
  • The geometric mean of $\frac{1}{1},\frac{1}{2},\ldots,\frac{1}{n}$ is bounded by the arithmetic mean of the same terms, which is $\frac{H_n}{n}$ and behaves like $\frac{\log(n)}{n}$. – Jack D'Aurizio Jan 08 '20 at 22:05

2 Answers2

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You can bound $n! \ge (n/2)^{n/2}$.

Then $(1/n!)^{1/n} \le (n/2)^{-1/2}$ which clearly goes to 0. Since $1/n!\ge 0$ the sandwich theorem gives the result.

Thomas Ahle
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You have $$ \begin{split} \lim_{n \to \infty} (n!)^{1/n} &= \exp\left( \lim_{n \to \infty} \ln\left((n!)^{1/n}\right) \right)\\ &= \exp\left( \lim_{n \to \infty} \frac{\ln(n!)}{n} \right) \\ &\ge \exp\left( \lim_{n \to \infty} \frac{\frac{n}{2} \ln(n/2)}{n} \right) \\ &\to \infty \end{split} $$

The main idea is to take logs to reduce the power and to use the fact that $$\ln(n!) = \sum_{k=1}^n \ln(k)$$ has an linear in $n$ amount of terms, which allows to cancel the $n$ in the denominator...

Your main issue was ignoring that the amount of terms is growing linearly with $n$

gt6989b
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    I think this can be simplified by saying $n! \ge (n/2)^{n/2}$ at the beginning and not bothering with the exponential and logarithm. – Trevor Gunn Jan 08 '20 at 20:20
  • @TrevorGunn +1, agree -- for this problem a better approach. But general approach to these problems (which I would like the OP to learn) is usually the technique in my answer. – gt6989b Jan 08 '20 at 20:23