A majoration of the spectral radius

Idea. The spectral radius is controlled by the norm of *every* power: $\rho(A)\le\|A^k\|^{1/k}$. For symmetric matrices the inequality is an equality, because the operator norm of a symmetric matrix is its spectral radius.

Step 1: eigenvalues of $A^k$. Triangularise $A$ over $\mathbb{C}$: $A=PTP^{-1}$ with $T$ upper triangular with diagonal $\lambda_1,\dots,\lambda_n$. Then $A^k=PT^kP^{-1}$ and $T^k$ is upper triangular with diagonal $\lambda_1^k,\dots,\lambda_n^k$. Hence the eigenvalues of $A^k$ are $\lambda_i^k$ and $\rho(A^k)=\rho(A)^k$.

Step 2: the inequality. By the problem "Eigenvalues are bounded by the operator norm" applied to $A^k$,
$$\rho(A)^k=\rho(A^k)\le\|A^k\|,$$
so $\rho(A)\le\|A^k\|^{1/k}$.

Step 3: symmetric matrices. If $A$ is symmetric, so is $A^k$, with eigenvalues $\lambda_i^k$. For symmetric matrices the operator norm is the largest absolute eigenvalue, so
$$\|A^k\|=\max_i|\lambda_i|^k=\rho(A)^k.$$
Hence $\|A^k\|^{1/k}=\rho(A)$ for every $k$, and the limit is $\rho(A)$ trivially.

Counter-example for non-symmetric matrices. For $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, $\rho(A)=1$, and $\|A^k\|^{1/k}$ equals $2.41,1.59,1.10,1.014$ for $k=1,5,50,500$: it decreases to $1$ but is $>1$ for each finite $k$.

Remark. The limit statement $\|A^k\|^{1/k}\to\rho(A)$ holds for *every* matrix (Gelfand's formula), not only symmetric ones; the proof for general $A$ requires the Jordan form or complex analysis (the resolvent). In particular, $\rho(A)<1$ iff $A^k\to0$.