 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/18/a/ii/solution

x_{i} = (1 + ϵ)^{i}

i = 0, 1, \dots

. Then

Δ x_{i} = ϵ (1 + ϵ)^{i}

, and its left-endpoint contribution is

f (x_{i}) Δ x_{i} = (1 + ϵ)^{- (1 + α) i} ϵ (1 + ϵ)^{i} = ϵ [(1 + ϵ)^{- α}]^{i} .

first term ϵ, q = (1 + ϵ)^{- α}, U_{ϵ} = \frac{ϵ}{1 - ( 1 + ϵ ) ^{- α}} .

Since

ϵ > 0

and

α > 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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