Theorem 3.37 in Baby Rudin: $\lim\inf\frac{c_{n+1}}{c_n}\leq\lim\inf\sqrt[n]{c_n}\leq\lim\sup\sqrt[n]{c_n}\leq \lim\sup\frac{c_{n+1}}{c_n}$

Here's Theorem 3.37 in the book Principles of Mathematical Analysis by Walter Rudin, third edition:

For any sequence $\{c_n\}$ of positive numbers, $$\lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n} \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ $$\lim_{n\to\infty} \sup \sqrt[n]{c_n} \leq \lim_{n\to\infty} \sup \frac{c_{n+1}}{c_n}.$$

Now Rudin has given a proof of the second inequality. Here's my proof of the first.

Let $$\alpha = \lim_{n\to\infty} \inf \frac{c_{n+1}}{c_n}.$$ Then $\alpha \geq 0$. If $\alpha = 0$, then we're done since $$\lim_{n\to\infty} \sqrt[n]{c_n} \geq 0.$$ So we suppose that $\alpha > 0$ and choose a real number $\beta$ such that $0 < \beta < \alpha$. Then by the result analogous to Theorem 3.17 (b), there is an integer $N$ such that $n \geq N$ implies $$\frac{c_{n+1}}{c_n} > \beta,$$ which in turn implies $$c_{n+1} > \beta c_n.$$ So for each $n \geq N$, we have $$c_n \geq \left( c_N \cdot \beta^{-N} \right) \cdot \beta^n.$$ Thus, for $n \geq N$, we have $$\sqrt[n]{c_n} \geq \beta \sqrt[n]{ c_N \cdot \beta^{-N} }.$$

Then taking the limit inferior of both sides, we get $$\lim_{n\to\infty}\inf \sqrt[n]{c_n} \geq \lim_{n\to\infty} \inf \left( \beta \sqrt[n]{ c_N \beta^N} \right) = \lim_{n\to\infty} \left( \beta \sqrt[n]{ c_N \beta^N} \right) = \beta \cdot 1 = \beta$$ by Theorem 3.20 (b).

Thus we have shown that for any (positive) real number $\beta$ such that $\beta < \alpha$, we have $$\beta \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ which implies that $$\alpha \leq \lim_{n\to\infty} \inf \sqrt[n]{c_n},$$ as required.

Is the above proof correct? If so, then is my presentation good enough? If not, then where does the problem lie?

• Correct, and your presentation is excellent. Apr 29 '16 at 14:14
• I concur with @UmbertoP. Apr 29 '16 at 14:14
• @Math1000 thanks. Have you changed anything in my post? What have you edited in it? Apr 29 '16 at 14:18
• Just a couple of spelling errors, nothing mathematically related. Apr 29 '16 at 14:18

I think there is a bit that is probably not quite correct both in the above proof and in Rudin's 3.37 due to the sign in inequalities with $\beta$. It should be either 'strictly less than' in 3.37 in the book, or 'strictly greater than' in the proof above.

To see that all inequalities involving $\beta$'s should all have strict signs in inequalities you may refer to Theorem 3.17

\begin{align} s_n < x \quad \text{ for } \frac{c_{n+1}}{c_n} < \beta \end{align}

Please, look page 66, proofs 3.33 and 3.34 top to bottom and page 67 very top. Why this is important? I will explain using the book's Theorem 3.37. Basically, we have a sequence with $\lim \sup$ equals $\alpha$, which means that if we choose any $\beta : \alpha < \beta$, then beyond some point all terms will be strictly less than $\beta$.

In a series of steps in the proof and using the same sequence $\{ c_n \}$ we show that

\begin{align} \lim \sup_{n \to \infty} \sqrt[n]{c_n} < \beta \end{align}

Finally, as we have chosen $\beta$ arbitrarily, $\lim \sup_{n \to \infty} \sqrt[n]{c_n} < \beta$ and $\alpha < \beta$, we may conclude because intuitively since by our arbitrary choice $\beta \in ( \alpha, \infty)$, as $\lim \sup_{n \to \infty} \sqrt[n]{c_n}$ is strictly less any $\beta$, it cannot lie in the segment $( \alpha, \infty)$ Therefore we have to put

\begin{align} \lim \sup_{n \to \infty} \sqrt[n]{c_n} \le \alpha = \lim \sup_{n \to \infty} \frac{c_{n+1}}{c_n} \end{align}

P.S

To fix the proof above (typos), we need to put $>$ above in \begin{align} c_n > \left( c_N \cdot \beta^{-N} \right) \cdot \beta^n \end{align} and down below, in all inequalities with $\beta$ !

P.P.S.

In addition to the above, it seems, the proof in Rudin's 3.37 contains totally irrelevant block starting from 'In particular' and ending with 'or' on page 68. There seems to be no link of this block with the rest of the proof. But who cares?

• I think that proof by Rudin is also correct. Because $a\lt b \implies a\le b$.
– Koro
Feb 23 '21 at 1:42