= Solution
The spectral expansion gives
$$
\left(\frac H{\lambda_n}\right)^ke_1
=\sum_{j=1}^n b_j\left(\frac{\lambda_j}{\lambda_n}\right)^kw_j.
$$
For $j\leq n-2$, the coefficient tends to zero because $|\lambda_j|<|\lambda_n|$. Thus every possible <limit> belongs to
$$
W=\operatorname{span}\{w_{n-1},w_n\}.
$$
The identical argument with the coefficients $c_j$ proves
$$
\boxed{v_1,v_2\in W.}
$$
The statement is conditional because, when $\lambda_{n-1}=-\lambda_n$, the component along $w_{n-1}$ generally alternates rather than converges.
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