Error in Spivak's proof that an odd degree polynomial has a root? I'm reading Spivak's Calculus 3rd edition. On page 123 (chapter 7, Three Hard Theorems) he gives a proof of the theorem that states if $n$ is odd, then any equation $x^n+a_{n-1}x^{n-1}+...+a_0=0$ has a root.
Essentially he's showing that there's an interval $[a, b]$ such that $f(a)<0<f(b)$ where  $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$, concluding that there's $x \in [a,b]$ such that $f(x)=0$.
The proof begins as follows:
$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right)$$
$$\left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\frac{|a_{n-1}|}{|x|}+...+\frac{|a_0|}{|x^n|}$$
If we choose $x$ satisfying
$$|x|>1, 2n|a_{n-1}|,...,2n|a_0|$$
then
$$\mbox{ (1) } |x^k|>|x|$$
and
$$\mbox{ (2) }\frac{|a_{n-k}|}{|x^k|}<\frac{|a_{n-k}|}{|x|}<\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$
Now he doesn't say anything about $k$. Since he stated that $|x^k|>|x|$, I assumed that $k\in[2,n]$ because if $k$ were in $[1,n]$ then it wouldn't have been true that $|x^k|>|x|$ - we'd have had $|x^1|>|x|$ which is false. But if $k\in[2,n]$, how does it follow that $\frac{|a_{n-1}|}{|x|}<\frac1{2n}$? If this doesn't, does (3) still holds?
In the next step he says that
$$\mbox{ (3) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$
where "less than or equals" doesn't really make sense - it should have been "strictly less" since we have "strictly less" in (2).
But everything would've made sense if he did the following:
$$\mbox{ (1*) } |x^k|\geq|x|, k\in [1,n]$$
$$\mbox{ (2*) }\frac{|a_{n-k}|}{|x^k|}\leq\frac{|a_{n-k}|}{|x|}\leq\frac{|a_{n-k}|}{2n|a_{n-k}|}=\frac1{2n}$$
$$\mbox{ (3*) } \left|\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right|\leq\sum_{1}^n\frac1{2n}=\frac12$$
Am I missing something, or it's indeed an error in the book?
 A: You are correct that
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}<\frac{\lvert a_{n-k}\rvert}{\lvert x\rvert}<\frac{\lvert a_{n-k}\rvert}{2n\lvert a_{n-k}\rvert}=\frac{1}{2n}
$$
only holds for $k\ge2$. However, for $k=1$ we have
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}\le\frac{\lvert a_{n-k}\rvert}{\lvert x\rvert}<\frac{\lvert a_{n-k}\rvert}{2n\lvert a_{n-k}\rvert}=\frac{1}{2n},
$$
where the strict part of the inequality holds by the assumption $\lvert x\rvert>2n\lvert a_{n-1}\rvert$. So for all $k\ge1$ the inequality
$$
\frac{\lvert a_{n-k}\rvert}{\lvert x^k\rvert}<\frac{1}{2n}
$$
holds. From this you can conclude that
$$
\left\lvert\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}\right\lvert<\sum_1^n\frac{1}{2n}=\frac{1}{2}.
$$
That he writes "$\le$" here instead of "$<$" does not make his inequality false, only weaker than it could be (and potentially confusing to readers). In the remainder of the proof he doesn't need the strict form of this inequality; the non-strict form he writes suffices.
A: You are right when you state

But everything would've made sense if he did the following: ....

Note that your approach requires the general assumption that $x$ is chosen such that
$$(0^*) \quad |x| \ge 1, 2n|a_{n-1}|,...,2n|a_0| .$$
Spivak's proof is nevertheless correct although it has little shortcomings. In fact, $(1)$ is true only for $k \ge 2$, thus also $(2)$ is true only for $k \ge 2$. However, the essence of $(2)$ is
$$(2') \quad \frac{|a_{n-k}|}{|x^k|} < \frac1{2n} ,$$
the intermediate inequalities are irrelevant. And this is true also for $k = 1$ because $|x| > 2n|a_{n-1}|$. The condition $|x| > 1$ is only needed to obtain $(1)$, but in fact it would also suffice to require  $|x| \ge 1$ to obtain
$$(2'') \quad \frac{|a_{n-k}|}{|x^k|} \le \frac{|a_{n-k}|}{|x|} < \frac{|a_{n-k}|}{2n|a_{n-k}|} = \frac1{2n} .$$
Anyway, in my opinion it is unnecessary to work with $(1)$ and $(2)$. We have
$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n}\right)$$
and clearly
$$1+\frac{a_{n-1}}{x}+...+\frac{a_0}{x^n} \to 1$$
as $\lvert x \rvert \to \infty$. Thus
$$f(x) \to \pm \infty$$
as $x \to \pm \infty$.
Therefore we find $a < 0 < b$ such that $f(a) < 0 < f(b)$.
