Contradiction in radii of convergence? Where is my error? I'm working through Baby Rudin and I came across what seems to me to be a contradiction, but it could be an error on my part.  It has to do with radii of convergence of power series.
First, let $\{a_n\}$ be a sequence of positive real numbers such that 
$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}=\alpha<1\tag{1}$$
and
$$\limsup_{n\to\infty}\sqrt[n]{a_n}=\beta<1.\tag{2}$$
By Theorem 3.37, $\beta\leq\alpha.$
Suppose $\beta<\alpha$ and thus $$\frac{1}{\alpha}<\frac{1}{\beta}.\tag{3}$$
Now consider the power series $S(x)=\sum_{n=0}^\infty{a_nx^n}.$
 By the root test and $(2)$, $S(x)$ converges when $|x|<\frac{1}{\beta}.$  And by the ratio test and $(1)$, $S(x)$ diverges when $|x|>\frac{1}{\alpha}.$
Now Choose a $x_o$ such that $$\frac{1}{\alpha}<x_0<\frac{1}{\beta}.\tag{4}$$ 
By the root test $S(x_0)$ converges.  By the ratio test $S(x_0)$ diverges. Contradiction.
Did I make an error, or is this a proof that $\beta=\alpha?$
 A: The issue is that the ratio test doesn't prove that the series in question diverges; that's in the inconclusive range. A corrected statement of the ratio test:
Let $c_n$ be a series of positive numbers. The sum $\sum_n c_n$ converges if $\limsup \frac{c_{n+1}}{c_n} < 1$. The sum $\sum_n c_n$ diverges if $\liminf \frac{c_{n+1}}{c_n} > 1$.
If $\limsup \frac{c_n+1}{c_n} \ge 1$ and $\liminf \frac{c_{n+1}}{c_n} \le 1$, the ratio test is inconclusive. For series with mixed signs, take absolute values of everything; this is a test for absolute convergence.
And now, an example: $f(x)=1+x+2x^2+2x^3+4x^4+4x^5+8x^6+8x^7+\cdots$; the coefficient $a_n$ of $x^n$ is $2^{\lfloor n/2\rfloor}$. The $\limsup$ of the ratio $\frac{a_{n+1}}{a_n}$ is $2$ and the $\liminf$ of the ratio is $1$. The $\limsup$ of the root $a_n^{1/n}$ is $\sqrt{2}$. The series converges, with exact value $\frac{1+x}{1-2x^2}$, for $|x| < \frac1{\sqrt{2}}$, exactly as the root test says. On the other hand, all the ratio test can tell us is that the series converges for $|x|<\frac12$ and diverges for $|x|>1$.
From a logical perspective, the ratio test is strictly weaker than the root test. The reason to use it is that it's often easier to find the ratio than the root.
