Examples 3.35 (a) and (b) in Baby Rudin: Limit Superior and limit inferior of a couple of sequences

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

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

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:

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

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