= Solution
The <binomial series> gives $(1+\epsilon)^{-\alpha}=1-\alpha\epsilon+O(\epsilon^2)$. Therefore
$$
U_\epsilon=\frac{\epsilon}{\alpha\epsilon+O(\epsilon^2)}\longrightarrow\frac1\alpha.
$$
To justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_\epsilon=(1+\epsilon)^{-1-\alpha}U_\epsilon$, and monotonicity gives $L_\epsilon\leq I\leq U_\epsilon$. Both bounds have the same limit. The <geometric partition bounds for a power-law improper integral> and the <squeeze theorem> thus prove
$$
\boxed{I=\frac1\alpha,\qquad\alpha>0.}
$$
This bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $\epsilon$.
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