= Solution
For a fixed $n$, convergence for every initial error is equivalent to $|1-\omega t_k|<1$ for all $k$, where $t_k=1-\cos(k\pi/(n+1))>0$. Thus
$$
0<\omega<\frac{2}{1+\cos(\pi/(n+1))}
=\frac1{\cos^2(\pi/[2(n+1)])}.
$$
A single parameter works for every finite $n$ precisely when \b[$\boxed{0<\omega\leq1}$]: the largest $t_k$ tends to two as $n\to\infty$, so every $\omega>1$ eventually fails, whereas $\omega=1$ has all <eigenvalues> strictly between $-1$ and one for each finite grid.
For such $\omega$ the lowest-frequency <eigenvalue> satisfies, using $1-\cos x\leq x^2/2$,
$$
\lambda_1(\omega)=1-\omega(1-\cos(\pi/(n+1)))
\geq1-\frac{\pi^2}{2(n+1)^2}.
$$
An initial error proportional to $v_1$ retains exactly the factor $\lambda_1^\nu$. Hence the worst-case <spectral radius> cannot improve on a factor of the form \b[$(1-c/n^2)^\nu$], with $c$ independent of grid size, and reducing a slow-mode error by a fixed fraction takes order $n^2$ iterations. This is a worst-case rate assertion, not a lower bound for every individual initial error: a vector containing only faster modes, or the zero vector, need not exhibit the slow mode.
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