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This question is related to Examples 3.35 (a) and (b) in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition, p. 67.

Let us consider the series $$ \frac 1 2 + \frac 1 3 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots$$ for which the formula for the general term $a_n$ is given by $$a_n = \begin{cases} \frac{1}{2^k} \ \mbox{ if } \ n = 2k-1 \\ \frac{1}{3^k} \ \mbox{ if } \ n = 2k \end{cases} $$ for $k = 1, 2, 3, \ldots$, and the series $$\frac 1 2 + 1 + \frac 1 8 + \frac 1 4 + \frac{1}{32} + \frac{1}{16} + \cdots$$ for which the formula for the general term is given by $$b_n = \begin{cases} \frac{1}{2^n} = \frac{1}{2^{2k-1}} \ \mbox{ if } \ n = 2k-1 \\ \frac{1}{2^{n-2}} = \frac{1}{2^{2k-2}} \ \mbox{ if } \ n = 2k \end{cases} $$ for $k = 1, 2, 3, \ldots$.

Now Rudin states that $$\liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left( \frac 2 3 \right)^n = 0,$$ $$\liminf_{n\to\infty} \sqrt[n]{a_n} = \lim_{n\to\infty} \sqrt[2n]{\frac{1}{3^n}} = \frac{1}{\sqrt{3}},$$ $$\limsup_{n\to\infty} \sqrt[n]{a_n} = \sqrt[2n]{\frac{1}{2^n}} = \frac{1}{\sqrt{2}},$$ $$\limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac 1 2 \left( \frac 3 2 \right)^n = +\infty.$$ And, Rudin also states that $$ \liminf_{n\to\infty} \frac{b_{n+1}}{b_n} = \frac 1 8,$$ $$ \limsup_{n\to\infty} \frac{b_{n+1}}{b_n} = 2,$$ $$\lim_{n\to\infty} \sqrt[n]{b_n} = \frac 1 2.$$

How to rigorously verify these statements using machinery (i.e. the definitions and theorems ) developed by Rudin up to this point?

I know that $\liminf$ and $\limsup$ are the infimum and supremum, resp., of the set of all the subsequential limits (in the extended real number system) of a sequence, and there is a subsequence each converging to $\liminf$ and $\limsup$.

Moreover, for each $k \in \mathbb{N}$, we have $$\frac{a_{2k} }{a_{2k-1}} = \frac{ \frac{1}{3^k} }{ \frac{1}{2^k} } = \left( \frac 2 3 \right)^k \to 0 \ \mbox{ as } \ k \to \infty,$$ $$\frac{ a_{2k+1} }{ a_{2k} } = \frac{ \frac{1}{2^{k+1}}}{ \frac{1}{3^k} } =\frac 1 2 \left( \frac 3 2 \right)^k \to +\infty \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k]{a_{2k}} = \sqrt[2k]{ \frac{1}{3^k}} = \sqrt{\frac 1 3} \to \sqrt{\frac 1 3} \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k-1]{a_{2k-1}} = \sqrt[2k-1]{ \frac{1}{2^k}} = \frac{1}{2^{\frac{k}{2k-1}}} \to ? \ \mbox{ as } \ k \to \infty.$$ How to find $\lim_{k \to \infty} \sqrt[2k-1]{2^k} $ using what Rudin has established (in Theorem 3.20)?

Also, $$\frac{ b_{2k} }{ b_{2k-1} } = \frac{ \frac{1}{ 2^{2k-2} } }{ \frac{1}{ 2^{2k-1} } } = 2 \to 2 \ \mbox{ as } \ k \to \infty,$$ $$\frac{ b_{2k+1} }{ b_{2k} } = \frac{ \frac{1}{ 2^{2k+1} } }{ \frac{1}{ 2^{2k-2} } } = \frac 1 8 \to \frac 1 8 \ \mbox{ as } \ k \to \infty,$$

$$\sqrt[2k]{b_{2k}} = \sqrt[2k]{\frac{1}{2^{2k-2}}} = \frac{ \sqrt[2k]{4} }{ 2 } \to \frac 1 2 \ \mbox{ as } \ k \to \infty,$$ $$\sqrt[2k-1]{b_{2k-1}} = \sqrt[2k-1]{\frac{1}{2^{2k-1}}} = \frac 1 2 \to \frac 1 2 \ \mbox{ as } \ k \to \infty,$$

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I will deal only with 3.35 (a) here.

Part I

Starting where you left off, I will prove that as $ k \to \infty $:

$$ \sqrt[2k-1]{\frac{1}{2^k}} \to \frac{1}{\sqrt{2}} $$

Consider sequence:

$$ \sqrt[n]{\frac{1}{2^{\frac{n + 1}{2}}}} $$

Obviously previous sequence is a subsequence of this one (take $ n = 2k-1 $).

We have:

$$ \sqrt[n]{\frac{1}{2^{\frac{n + 1}{2}}}} = \sqrt[n]{\frac{1}{2^{\frac{n}{2}}\sqrt{2}}} = \frac{1}{\sqrt{2}\sqrt[n]{\sqrt{2}}} $$

Since $ \sqrt{2} > 0 $, we have $ \sqrt[n]{\sqrt{2}} \to 1 $ as $ n \to \infty $ (Rudin 3.20 (b)).

Thus $$ \sqrt[n]{\frac{1}{2^{\frac{n + 1}{2}}}} \to \frac{1}{\sqrt{2}} $$

and thus (by note after definition 3.5 in Rudin) its subsequence has to converge to the same number, so:

$$ \sqrt[2k-1]{\frac{1}{2^k}} \to \frac{1}{\sqrt{2}} $$

Part II

Let's deal with a sequence $ \frac{a_{n+1}}{a_n} $. Let $ E $ be the set of all subsequential limits of this sequence (including potenatially $ -\infty $ and $ + \infty $).

It was shown in question that $ + \infty, 0 \in E $.

Thus $ \sup E = + \infty $, since for $ e \in E $ we obviously have $ e \leq + \infty $ and for any $ x < +\infty $ $ x $ is not an upper bound on $ E $, since otherwise we would have $ + \infty \leq x $ ($ + \infty $ is in $ E $!).

Note also that for every $ n $ we have $ \frac{a_{n+1}}{a_n} > 0 $ and thus any for any limit $ e \in E $ we have $ e \geq 0 $, thus $ \inf E = 0 $, since if $ x > 0 $ it can't be a lower bound ($ 0 $ is in $ E $).

This by definition means:

$$ \lim_{n \to \infty}\sup \frac{a_{n+1}}{a_n} = + \infty $$ $$ \lim_{n \to \infty}\inf \frac{a_{n+1}}{a_n} = 0 $$

Part III

Let's deal now with $ \sqrt[n]{a_n} $.

It was shown in the question that $ \frac{1}{\sqrt{3}} \in E $. I have shown that $ \frac{1}{\sqrt{2}} \in E $

Having theorem 3.17 in Rudin in mind, assume:

$$ x > \frac{1}{\sqrt{2}} $$

We will do proof by contradiction. Assume that for any $ n_0 $ exists $ n \geq n_0 $ such that:

$$ \sqrt[n]{a_n} > x > \frac{1}{\sqrt{2}} $$

Depending on $ n $ being odd or even, there are two cases:

$$ \sqrt[2k]{\frac{1}{3^k}} > x > \frac{1}{\sqrt{2}} $$

and thus

$$ \frac{1}{3^k} > (\frac{1}{\sqrt{2}})^{2k} = \frac{1}{2^k} $$

which is impossible.

In second case:

$$ \sqrt[2k - 1]{\frac{1}{2^k}} > x > \frac{1}{\sqrt{2}} $$

and thus

$$ \frac{1}{2^k} > (\frac{1}{\sqrt{2}})^{2k - 1} = \frac{1}{2^{\frac{2k-1}{2}}} = \frac{1}{2^k}\sqrt{2} $$

and thus $ 1 > \sqrt{2} $ which is false.

Thus by contradiction we have shown that

$$ \lim_{n \to \infty}\sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $$

Similarily you would show:

$$ \lim_{n \to \infty}\inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $$

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  • $\begingroup$ For the first part, perhaps it is simpler to observe that $ (2^{-k})^{1/(2k-1)}=$ $2^{-k/(2k-1)}=$ $2^{-1/2}\cdot (1/2^{1/(4n-2)}).$ $\endgroup$ – DanielWainfleet Jul 16 '18 at 0:56

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