Spectral Radius Formula 
An element of a Banach algebra $\frak{U}$ has spectral radius $r(A)$ given by the formula $$r(A) = \lim_{n \to \infty} ||A^n||^{1/n}$$. In particular, $\{||A^n||^{1/n}\}$ has a limit. 

Here is the proof from Kadison and Ringrose's book on Operator Algebras:

The part I am having trouble is with the sentence "Thus, if $0 \le a' < \overline{\lim} ||A^n||^{1/n}$..." Why does there exist such an $a$? And why does $a'$ being an arbitrary non-negative number less than...imply that inequality (7) holds. I'm drawing a blank for some reason. 
 A: For your first question: note that they say

From Theorem 3.3.1... this series represents $f$ on the largest open disk with center $0$ on which $f$ is defined.

As stated earlier, $f$ refers to the function $f(z) = (I - zA)^{-1}$. Now, the sentence before the conclusion we want to understand says:

On the other hand, Theorem 3.3.2 informs us that this series fails to converge for $z$ of modulus exceeding $(\overline{\lim}\|A^n\|^{1/n})^{-1}$.

In other words: if $|z| > (\overline{\lim}\|A^n\|^{1/n})^{-1}$, then $z$ lies outside of the open disk that is referred to in my first quote. In other words, for any $z' > (\overline{\lim}\|A^n\|^{1/n})^{-1}$, there necessarily exists a complex number $w$ with $|w| < z'$ for which $f(w)$ is not defined.
Now, if we take $a' = 1/z'$ and $a = 1/w$, then we can say that for any $0\leq a' < \overline{\lim}\|A^n\|^{1/n}$, there exists an $a$ with $|a| > a'$ for which $f(a^{-1})$ is note defined. Note that
$$
f(a^{-1}) = (I - a^{-1}A)^{-1},
$$
so this is exactly the statement we wanted to understand.

For your second question: putting the third-to-last and second-to-last sentences together tells us that whenever $0 \leq a' < \overline{\lim}\|A^n\|^{1/n}$, we must have $a' < r(A)$.  That is, 
$$
[0,\overline{\lim}\|A^n\|^{1/n}) \subseteq [0,r(A)).
$$
Of course, this means that $\overline{\lim}\|A^n\|^{1/n} \leq r(A)$.
