= Solution
Index the stated <geometric progression> by $x_i=(1+\epsilon)^i$, $i=0,1,\ldots$. Then $\Delta x_i=\epsilon(1+\epsilon)^i$, and its left-endpoint contribution is
$$
f(x_i)\Delta x_i=(1+\epsilon)^{-(1+\alpha)i}\epsilon(1+\epsilon)^i
=\epsilon\big[(1+\epsilon)^{-\alpha}\big]^i.
$$
Hence the approximation is a <geometric series> with
$$
\boxed{\text{first term }\epsilon,\qquad q=(1+\epsilon)^{-\alpha},\qquad
U_\epsilon=\frac{\epsilon}{1-(1+\epsilon)^{-\alpha}}.}
$$
Since $\epsilon>0$ and $\alpha>0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the <improper integral>.
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