# Proving a lim sup inequality in Chapter 12 of Kenneth Ross's “Elementary Analysis”

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!

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.

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.

• Thank you for your response. I can follow your proof up until the conclusion, but I still cannot see how we can prove that α ≤ L by showing that α ≤ β for any β > L. If we let β=L+ϵ where ϵ>0, then L<L+ϵ. Now with α ≤ L+ϵ, α=L+ϵ is permitted, but this implies that α>L which is the opposite of what we want...what am I missing here? – user84815 Jul 26 '13 at 18:58
• @user84815 I am not sure what you mean. I my proof, I prove the contrapositive of the statement, that is, instead of proving $p\to q$ and $q\to p$ y prove $\neg q\to\neg p$ and $\neg p\to\neg q$. – Pedro Tamaroff Jul 26 '13 at 19:51
• You proved that ℓ ≤ β (let β=lim inf a(n)^(1/n)) by showing that ℓ-ϵ≤β where ℓ-ϵ<ℓ. But if β=ℓ-ϵ (which is permitted by ≤) then β<ℓ, which violates ℓ ≤ β...the condition we are trying to prove. In other words, how does the second to last inequality in your proof imply the final inequality in your proof? Thanks again for the help. – user84815 Jul 26 '13 at 20:23
• Or am I supposed to assume that because ϵ can be made arbitrarily small that ℓ-ϵ≤β for all ℓ-ϵ implies ℓ≤β? This assumption, if valid, still confuses me inasmuch ϵ>0, even for arbitrarily small values of ϵ. – user84815 Jul 26 '13 at 22:36
• @user84815 Sorry, I was busy. You're mixing up $\ell$s, it seems. I chose $\ell$ to denote a limit, and then I had denoted it for some arbitrary numbers. I'm sorry if it confused you. – Pedro Tamaroff Jul 26 '13 at 23:09

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 $b=L$.