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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)?

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  • $\begingroup$ @Dr.MV yes, you're right. I have already stated Theorem 3.39. Would you like me to also include Theorems 3.33 and 3.34? $\endgroup$ Apr 29 '16 at 15:00
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    $\begingroup$ Shouldn't be necessary. You said what these theorems are, that suffices, they're well-known enough. You can get $\limsup \dfrac{1}{\sqrt[n]{n!}} = 0$ from the fact that the exponential series has infinite radius of convergence, which you can easily get from the ratio test. $\endgroup$ Apr 29 '16 at 15:03
  • $\begingroup$ If I infer correctly the meaning of your last question, then, yes, the ratio test does apply for power series with complex summands. $\endgroup$
    – Mark Viola
    Apr 29 '16 at 15:26
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There are at least $n/2$ numbers in $\{1,\ldots,n\}$ that are greater than or equal to $\frac{n}2$, so $$n! \ge \left(\frac{n}2\right)^{n/2} \implies (n!)^{1/n} \ge \sqrt{\frac{n}2} \to \infty.$$

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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 $m$.

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  • $\begingroup$ how to show this? Using induction on $m$, the result holds for $m=0$ (we can take $N=1$) and $m=1$ (we take $N=2$). Supposing the reslut holds for all $k = 0, 1, \ldots, m$ (with $N_0, N_1, \ldots N_m$), we take $N > \max( N_0, \ldots, N_m)$ and note that $$(m+1)^N = \sum_{k=0}^N {N \choose k} m^k \leq \sum_{k=0}^N {N \choose k} m^N = (2m)^N. $$ What next? $\endgroup$ Apr 29 '16 at 14:55
  • $\begingroup$ @SaaqibMahmuud Induction won't help much here. Fix an $m$ and use the fact that all but finitely many factors of $n!$ are $\ge 2m$. What happens as $n$ gets large? $\endgroup$
    – Erick Wong
    Apr 29 '16 at 15:13
  • $\begingroup$ @ErickWong can you please elaborate? Yes, for a given $m$, if we take $N > 4m$, say, then the first $2m-1$ factors in $N!$ are less than $2m$ and the remaining $2m+1$ characters at least are $\geq 2m$. So as $N$ gets larger and larger, the latter number gets more and more. What next? $\endgroup$ Apr 30 '16 at 5:21
  • $\begingroup$ @SaaqibMahmuud Use what you have to calculate a lower bound (it doesn't even have to be a good lower bound) for the ratio $N!/m^N$, then show that for any fixed $m$, this ratio becomes $>1$ for large enough $N$. $\endgroup$
    – Erick Wong
    Apr 30 '16 at 5:30
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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 $n\geq 2.$

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  • $\begingroup$ A rigorous Rudin-level treatment of Stirling's formula is very much overkill for this bound :). $\endgroup$
    – Erick Wong
    Apr 30 '16 at 4:35
  • $\begingroup$ @Eric Wong 8. That's why I only included an integral lower bound sufficient for the Q. $\endgroup$ Apr 30 '16 at 13:09
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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

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

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    $\begingroup$ the point is to obtain our desired within the context of Rudin's discussion. That's how rigorous mathematics works; you have to prove new results using the arsenal you've developed up to that point. $\endgroup$ Apr 30 '16 at 5:27
  • $\begingroup$ @SaaqibMahmuud I don't understand the context of the comment. Was there any tools used in this development that are beyond the level of the text? I have never read that particular text and so proceeded using elementary tools only. Please let me know how I can improve my answer. I really want to give you the best one I can. If you didn't find this approach useful, I am happy to delete it. Just let me know, OK? -Mark $\endgroup$
    – Mark Viola
    Apr 30 '16 at 15:15
  • $\begingroup$ @Dr.MV I wasn't familiar with the sequencing of baby Rudin either, so I looked it up: maa.org/press/maa-reviews/principles-of-mathematical-analysis. Sequences and series are developed (rigorously) in Chapter 3, differentiation in Chapter 5 and Riemann-Stieltjes integration in Chapter 6. $\endgroup$
    – Erick Wong
    Apr 30 '16 at 16:21

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