Example 3.40 (b) in Baby Rudin: How to find $\lim_{n\to\infty} \sup \frac{1}{\sqrt[n]{n!}}$?

Question

Here is Theorem 3.39 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:

Given the power seires $\sum c_n z^n$, put $$\alpha = \lim_{n\to\infty}\sup\sqrt[n]{\vert c_n \vert}, \ \ \ R = \frac{1}{\alpha}.$$ (If $\alpha = 0$, $R = +\infty$; if $\alpha = +\infty$, $R = 0$.) Then $\sum c_n z^n$ converges if $\vert z \vert < R$, and diverges if $\vert z \vert > 1$.

The proof makes use of the root test, which is Theorem 3.33. Now in Example 3.40 (b), Rudin states that the series $\sum {z^n \over n!}$ has $R = +\infty$, meaning $\alpha = 0$. How does this hold, especially in view of the fact that $\lim_{n\to\infty} \sqrt[n]{n} = \lim_{n\to\infty} \sqrt[n]{p} = 1$ for any $p > 0$, as has been stated and proved by Rudin in Theorem 3.20 (b) and (c)? That is, how to rigorously show (using only the machinery developed by Rudin until Theorem 3.39) that $$\lim_{n\to\infty} \sup {1 \over \sqrt[n]{n!}} = 0?$$

Can we state for the power series $\sum c_n z^n$ a result analogous to Theorem 3.39, using the ratio test (i.e. Theorem 3.34)?

Answer 1

Try proving that for every natural number $m\geq 1$, $n!$ is eventually larger than $m^n$ (in the sense that $n! > m^n$ for all $n>N$, for some $N$). This implies that $\limsup_n (n!)^{-1/n} < 1/m$ for all $m$.

Answer 2

There are at least $n/2$ numbers in $\{1,\ldots,n\}$ that are greater than or equal to $\frac{n}2$, so $$n! \ge \left(\frac{n}2\right)^{n/2} \implies (n!)^{1/n} \ge \sqrt{\frac{n}2} \to \infty.$$

Answer 3

For $n>1$ we have $(n!)^{-1/n}=$ $(\prod_{j=1}^n(1/j))^{1/n}<$ $(\sum_{j=1}^n (1/j))/n=$ $(1/n)O(\ln n)=o(1).$

You can also use Stirling's Formula : $n!=(1+d_n)(n/e)^n\sqrt {2 \pi n}$ where $|d_n|<1/6 n$ for $n\geq 1.$ For this Q, nothing as precise as this is needed: When $m\geq 2$ we have $\ln m>\int_{m-1}^m \ln x \;dx.\;$ So $\ln (n!)>\int_1^n\ln x\;dx= 1+n\ln n, $ for $n\geq 2.$

Answer 4

Note that

$$\sqrt[n]{n!}=e^{\frac1n \log(n!)} \tag 1$$

Next, we can write

$$\begin{align} \frac1n \log(n!)&=\frac1n \sum_{k=1}^n\log(k)\\\\ &=\log(n)+\frac1n \sum_{k=1}^n\log(k/n) \tag 2 \end{align}$$

Finally, note that

$$-1=\int_0^1 \log(x)\,dx\le \frac1n \sum_{k=1}^n\log(k/n)\le 0 \tag 3$$

Putting together $(1)-(3)$, we find that

$$\frac ne\le \sqrt[n]{n!}\le n$$

Since $n/e \to \infty$, we find the coveted limit

$$\lim_{n\to \infty}\sqrt[n]{n!}=\infty$$

Answer 5

Since Rudin gives us such a hint in the text: "In this case the ratio test is easier to apply than the root test." Let us try to activate the ratio test.

Proof: Put $a_{n}=\frac{1}{n!}\cdot z^{n}$, then apply the ratio test:

$$\limsup_{n \to \infty}|\frac{a_{n+1}}{a_{n}}| = \limsup_{n \to \infty}|\frac{z}{n+1}| = |z| \limsup_{n \to \infty}\frac{1}{n+1} = 0.$$

By Theorem 3.37 of Rudin's book, we have

$$\limsup_{n \to \infty} \sqrt[n]{|a_{n}|} \leq \limsup_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|.$$ Notice that $\{a_{n}\}$ is a sequence of positive numbers. Hence,

$$0 \leq \limsup_{n \to \infty} \sqrt[n]{|a_{n}|} \leq \limsup_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| = 0,$$ which means

$$\limsup_{n \to \infty} \sqrt[n]{|a_{n}|} = 0.$$ Therefore, $$R = \frac{1}{\alpha} = +\infty.$$

I took the inspiration from the proof of Theorem 3.39, and I think this is exactly what Rudin wants readers to do.