Geometric partition bounds for a power-law improper integral (${\int }_1^{\infty }{x}^{-1-\alpha }dx=1/\alpha ,\alpha >0$)

On the geometric progression $x_i=(1+ϵ{)}^i$, with $ϵ>0$, the interval widths are $ϵ(1+ϵ{)}^i$. The left-endpoint Riemann sum for the decreasing integrand $x^{-1-\alpha }$ is the geometric seriesThe right-endpoint sum is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$. Integral comparison on each interval gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds tend to $1/\alpha$ as $ϵ\to 0^{+}$, 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.