= Geometric partition bounds for a power-law improper integral
{title2=$\int_1^\infty x^{-1-\alpha}dx=1/\alpha,\quad\alpha>0$}
On the <geometric progression> $x_i=(1+\epsilon)^i$, with $\epsilon>0$, the interval widths are $\epsilon(1+\epsilon)^i$. The left-endpoint <Riemann sum> for the decreasing integrand $x^{-1-\alpha}$ is the <geometric series>
$$
U_\epsilon=\sum_{i=0}^\infty\epsilon(1+\epsilon)^{-\alpha i}
=\frac{\epsilon}{1-(1+\epsilon)^{-\alpha}}.
$$
The right-endpoint sum is $L_\epsilon=(1+\epsilon)^{-1-\alpha}U_\epsilon$. Integral comparison on each interval gives $L_\epsilon\leq I\leq U_\epsilon$. Both bounds tend to $1/\alpha$ as $\epsilon\to0^+$, proving the <improper integral> by the <squeeze theorem>. Although the largest interval width is unbounded on the infinite domain, these explicit upper and lower bounds justify the limit without incorrectly applying a finite-interval mesh argument to the entire half-line.
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