10. (a) Since $a_{n-1}1$, and let $a_{k}=q^{k}, k=1,2, \ldots$. Then $\left(1-a_{k-1} / a_{k}\right) / \sqrt{a_{k}}=(1-1 / q) / q^{k / 2}$, and consequently
$$b_{n}=\left(1-\frac{1}{q}\right) \sum_{k=1}^{n} \frac{1}{q^{k / 2}}=\frac{\sqrt{q}+1}{q}\left(1-\frac{1}{q^{n / 2}}\right) .$$
Since $(\sqrt{q}+1) / q$ can be arbitrarily close to 2, one can set $q$ such that $(\sqrt{q}+1) / q>b$. Then $b_{n} \geq b$ for all sufficiently large $n$. Second solution. (a) Note that
$$b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}=\sum_{k=1}^{n}\left(a_{k}-a_{k-1}\right) \cdot \frac{1}{a_{k}^{3 / 2}}$$
hence $b_{n}$ represents exactly the lower Darboux sum for the function $f(x)=x^{-3 / 2}$ on the interval $\left[a_{0}, a_{n}\right]$. Then $b_{n} \leq \int_{a_{0}}^{a_{n}} x^{-3 / 2} d x1$ such that $\int_{1}^{\alpha} x^{-3 / 2} d x>$ $b+(2-b) / 2$. Now, by Darboux's theorem, there exists an array $1=$ $a_{0} \leq a_{1} \leq \cdots \leq a_{n}=\alpha$ such that the corresponding Darboux sums are arbitrarily close to the value of the integral. In particular, there is an array $a_{0}, \ldots, a_{n}$ with $b_{n}>b$.