= Solution
Interpret boundedness above uniformly: there is an $M<\infty$ such that $B^k\preceq MI$ for every $k$. Since $B^k$ is a <positive-definite symmetric matrix>, the <spectral theorem> implies $(B^k)^{-1}\succeq M^{-1}I$. Therefore
$$
-\nabla f(x^k)^Ts^k=\nabla f(x^k)^T(B^k)^{-1}\nabla f(x^k)\geq\frac1M\|\nabla f(x^k)\|^2.
$$
Combining the <Armijo condition> with $\alpha^k\geq C>0$ yields
$$
f(x^k)-f(x^{k+1})\geq\frac{\beta_1C}{M}\|\nabla f(x^k)\|^2.
$$
If $f\geq L$, a <telescoping sum> gives
$$
\frac{\beta_1C}{M}\sum_{k=0}^N\|\nabla f(x^k)\|^2\leq f(x^0)-f(x^{N+1})\leq f(x^0)-L.
$$
The nonnegative <series> converges, so its terms tend to zero:
$$
\boxed{\lim_{k\to\infty}\|\nabla f(x^k)\|=0}.
$$
This proves first-order global convergence in the sense of <stationarity from uniformly bounded quasi-Newton matrices>. It does not imply that every nonconvex <objective function> reaches a <global minimum>, or that the iterates themselves converge. Every finite <limit point> is a <stationary point>, by <continuity> of the <gradient>.
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