Proving a lim sup inequality in Chapter 12 of Kenneth Ross's "Elementary Analysis"

Question

I am struggling to digest and understand Theorem 12.2 (p. 76) in Elementary Analysis: The Theory of Calculus by Kenneth Ross

Theorem 12.2

Let $s_n$ be any sequence of nonzero real numbers. Then we have

lim sup $\mid$$s_n$$\mid$$^($$^1$$^/$$^n$$^)$ $\leq$ lim sup $\mid$$s_n$$_+$$_1$/$s_n$$\mid$

In proving this, Ross suggests the following:

Let $\alpha$ = lim sup $\mid$$s_n$$\mid$$^($$^1$$^/$$^n$$^)$ and L = lim sup $\mid$$s_n$$_+$$_1$/$s_n$$\mid$. We need to prove that $\alpha$ $\leq$ L. This is obvious if L = + $\infty$, so we assume that L < + $\infty$. To prove $\alpha$ $\leq$ L it suffices to show:

$\alpha$ $\leq$ $\beta$ for any $\beta$> L

I'm curious as to why/how it suffices to show that $\alpha$ $\leq$ $\beta$ for any $\beta$ > L . How does this condition prevent $\alpha$ = $\beta$> L?

Thank you for your time and effort!

Answer 1

The reason is the following:

$$\alpha \leq \beta \iff \text{ for each }\epsilon >0 \text{ we have } \alpha< \beta+\epsilon$$

You can see with this is true, intuitively, by drawing the situation in the number line. This should help. One can of course give a proof, of the seemingly more obivous

$$x\leq 0\iff \text{ for each }\epsilon >0 \text{ we have } x<\epsilon$$

P If $x>0$, $x>x/2>0$ is true, with $\epsilon =x/2$. If for some $\epsilon>0$, $x\geq \epsilon >0$ then it follows $x>0$. Note we proved the contrapositives.

Now, since any number greater than $\alpha$ may be written as $\alpha+\epsilon$ for $\epsilon >0$, you get your claim.

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Set $a_n=|s_n|$. I will give you the proof for $\liminf$; and hope you can give that for $\limsup$. Note inequalities must be reversed! For example, you should choose $\alpha >\ell$.

P Set $$\ell =\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$$

Choose $\alpha <\ell$. By the definition of $\liminf$, there must exist an $N$ such that, for each $n\geq N$, we have that $$\alpha <\frac{a_{n+1}}{a_n}$$ That is, for $k\geq 0$, we have $$ a_{N+k}>\alpha\cdot a_{N+k-1}$$

This gives that $$a_{N+k}>\alpha^k \cdot a_{N}$$

In paricular $n-N\geq 0$ when $n=N+1,\dots$; so

$$a_{n}>\alpha^{n-N} \cdot a_{N}$$

Now, taking the $n$-th root gives that for $n\geq N+1 $ $$\root n \of {{a_n}} > \alpha \cdot{\left( {\frac{{{a_N}}}{{{\alpha ^N}}}} \right)^{1/n}}$$

and taking $\liminf\limits_{n\to\infty}$ gives

$$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $$

This means that for each $\alpha <\ell$,

$$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $$

which is saying that $$\ell \leq \liminf\limits_{n\to\infty} \root n \of {{a_n}}$$

The proof for $\limsup$ is completely analogous.

Answer 2

I'm assuming you're able to follow the rest of the proof given by Ross. There's a reason why he chose $\alpha $ and $ \beta$, mainly because it is relatively easier to find an upper bound to the ratio of a sequence, $L$, by some $\beta$. In the case you state, (suppose $\alpha = \beta$), we'd run into a contradiction, (this is shown towards the end of the proof where we show that if $\beta$ is an upper bound of $L$ then it must also be an upper bound on $\alpha$, which obviously wouldn't work if $\alpha = \beta$.)

But in all simplicity, refer to Exercise 3.8:

"Let $a,b \in \mathbb{R}$. Show if $a \leq b_1$ for every $b_1 >b$, then $a\leq b$."

This is exactly what Ross argues before starting his proof, just put $a=\alpha$, $b_1 = \beta$, and $b=L$.