# Spectral radius equal to 1 and convergence

The following theorem is well-known: $$\lim_k A^k = 0 \text{ if and only if } \rho(A)<1$$ (see wiki for context and proofs).

What if now $$\rho(A)=1$$ and $$\lambda\neq -1$$ for all $$\lambda \in Spec(A)$$. Then we also have convergence ? (not to $$0$$ but $$A$$ converges)

No. This already fails when all eigenvalues of $$A$$ are ones. E.g. $$A^k=\pmatrix{1&k\\ 0&1}$$ doesn't converge when $$A=\pmatrix{1&1\\ 0&1}$$.
When $$|\lambda|=1$$ but $$\lambda\ne1$$, we don't even have convergence of $$\lambda^k$$.