Shannon capacity definition I have seen the Shannon capacity defined in two ways:
$\Theta(G) = \sup_k \sqrt[k]{\alpha(G^k)}$
$\Theta(G) = \lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}$
My question is, basically, how do we know the limit (in the bottom one) exists, and how do we know the two definitions are equal?
$G^k$, by the way, stands for the strong product of $k$ copies of $G$.
I do know that $\alpha(G \boxtimes H) \geq \alpha(G) \cdot \alpha(H)$.  To see this, let $I_G$, $I_H$ be independent sets in $G$ and $H$.  Then, $I_G \times I_H$ is independent in $G \boxtimes H$.  At first, I mistakenly decided this meant the sequence $\sqrt[k]{\alpha(G^k)}$ must be nondecreasing.  From this, I made sense that the supremum and limit are the same.  Since an upper bound is $\chi(\bar{G})$, the limit definition exists.  But, this was a bad assumption.  I have discovered, using Sage, that
$\alpha(C_5) = 2$
$\sqrt{\alpha(C_5^2)} = \sqrt{5}$
$\sqrt[3]{\alpha(C_5^3)} = \sqrt[3]{10}$
And, $\sqrt{5} > \sqrt[3]{10}$.
 A: Without loss of generality, assume that the graph is not complete. Let $\alpha_n := \alpha(G^n)$; we need the following facts (noted by the OP in the question):


*

*$\alpha_n$ is montononically increasing.

*$\alpha_n \geqslant 2$.

*$\alpha_{n+m} \geqslant \alpha_n \alpha_m$.

*$\alpha_{nm} \geqslant \alpha_n^m$. (This is a corollary of item (3).)

*$\alpha_n \geqslant \alpha_1^n \geqslant 2^n$. 


Let $L$ be the supremum of $\alpha_n^{1/n}$ for all $n$. From (5.), it is clear that $L \geqslant 2$. We want to prove that $\alpha_n^{1/n}$ approaches $L$ as $n \to \infty$. 
Fix any $\varepsilon \in (0,1)$. Then there exists $k \in \mathbb N$ such that $\alpha_k \geqslant (L - \varepsilon)^k$. Now, for any $n$, define $q := \lfloor n/k \rfloor$. Then we have the following chain of inequalities:
$$
\begin{eqnarray*}
\alpha_n 
&\geqslant& 
\alpha_{kq} 
\\ &\geqslant& 
\alpha_k^q 
\\ &\geqslant& 
(L - \varepsilon)^{kq} 
\\ &\stackrel{(*)}{\geqslant}&
(L - \varepsilon)^{n - k} 
\\ &\geqslant&
\frac{(L - \varepsilon)^n}{L^{k}} 
\end{eqnarray*}
$$
As an exercise, justify the above steps carefully. [[The following facts are helpful to show $(\ast)$: (i) $kq \geqslant n - k + 1 \geqslant n - k$; and (ii) $L - \varepsilon \geqslant 2 - 1 = 1$.]]
From the chain of inequalities, we can conclude that
$$
\alpha_n^{1/n} \geqslant  \frac{L - \varepsilon}{L^{k/n}}.
$$
Now, taking $n$ sufficiently large, we can make the denominator at most, say, $1 + \varepsilon$. That is, there exists some $N$ such that for all $n \geqslant N$, we have $\alpha_n^{1/n} \geqslant \frac{L - \varepsilon}{1 + \varepsilon}$. Since $\varepsilon > 0$ is arbitrary, it follows that $\alpha_n^{1/n} \to L$ as $n \to \infty$.
