Yes, choose $a_1$ such that $\frac{1}{b_1}=tan(a_1)$, and let $a_{k+1}=a_k-arctan\frac{1}{k}$, then we have $\frac{1}{b_k}=tan(a_k)$, since $\sum_k arctan\frac{1}{k}=\infty$, the conjecture follows.

Yes, choose $a_1$ such that $\frac{1}{b_1}=\tan(a_1)$, and let $a_{k+1}=a_k-\arctan\frac{1}{k}$, then we have $\frac{1}{b_k}=\tan(a_k)$, since $\sum_k \arctan\frac{1}{k}=\infty$, the conjecture follows.