On the Shannon Capacity of Cycles
It is a standard fact that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$ for every graph $G$. (See the proof below.)
Denoting $b_k := \sqrt[k]{\alpha(G^k)}$, this implies that the tail of the sequence $(b_k)_{k \geq 1}$ cannot be monotone decreasing, and hence there are infinitely many $k$'s for which $b_{k+1} \geq b_k$, as required.



In order to prove that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$ note that the sequence $(\alpha(G^k))_{k \geq 1}$ is super multiplicative, i.e., $\alpha(G^{k+\ell}) \geq \alpha(G^k) \alpha(G^\ell)$ for all $k,\ell \geq 1$. Indeed, if $I$ is an independent set in $G^k$, and $J$ is an independent set in $G^\ell$, then $I \times J$ is an independent set in $G^{k+\ell}$. Therefore, by (the multiplicative version of) Fekete's lemma it follows that $\lim_{k \to \infty} \sqrt[k]{\alpha(G^k)}=\sup_{k \to \infty} \sqrt[k]{\alpha(G^k)}$.