A^k converges to 0 if and only if the spectral radius is less than 1

($\Rightarrow$). Suppose $\rho(A)\ge1$ and take an eigenvalue $\lambda\in\mathbb{C}$ with $|\lambda|\ge1$ and an eigenvector $v\in\mathbb{C}^n$, $v\ne0$. Then $A^kv=\lambda^kv$ and $|A^kv|=|\lambda|^k|v|\ge|v|>0$ for all $k$. Hence $A^kv\not\to0$, so $A^k\not\to0$. Contrapositive: $A^k\to0\Rightarrow\rho(A)<1$.

($\Leftarrow$). Assume $\rho(A)<1$. Write $A=PJP^{-1}$ in Jordan form over $\mathbb{C}$; then $A^k=PJ^kP^{-1}$, and it suffices to show $J^k\to0$, i.e. $J_m(\lambda)^k\to0$ for each block with $|\lambda|<1$. Write $J_m(\lambda)=\lambda I+N$ with $N^m=0$ and $\lambda I$ commuting with $N$. By the binomial theorem
$$J_m(\lambda)^k=\sum_{j=0}^{m-1}\binom kj\lambda^{k-j}N^j.$$
For fixed $j$, $\binom kj\le k^j$, so
$$\left|\binom kj\lambda^{k-j}\right|\le k^j|\lambda|^{k-j}\xrightarrow[k\to\infty]{}0,$$
because geometric decay beats polynomial growth ($|\lambda|<1$; if $\lambda=0$ the term vanishes for $k>j$). A finite sum of terms tending to $0$ tends to $0$.

Example. $A=\begin{pmatrix}0.9&1\\0&0.9\end{pmatrix}$: $A^k=\begin{pmatrix}0.9^k&k\,0.9^{k-1}\\0&0.9^k\end{pmatrix}$. The off-diagonal entry first grows (it is $\approx1.7\times10^{...}$, maximal near $k=9$) but eventually tends to $0$.

Remark. The transient growth is the reason why $\rho(A)<1$ does not imply $\|A\|<1$ for a given norm. A norm with $\|A\|<1$ exists iff $\rho(A)<1$ (see Gelfand's formula below).