 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2/10/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For

f (x) = x^{- 1 - p}

with

p > 0

[1, N]

gives

\sum_{n = 2}^{N} n^{- 1 - p} < I_{N} < \sum_{n = 1}^{N - 1} n^{- 1 - p}

, where

I_{N} = (1 - N^{- p}) / p

. Add the missing endpoint term on each side to obtain

\frac{1 - N ^{- p}}{p} + N^{- 1 - p} < n = 1 \sum N n^{- 1 - p} < \frac{1 - N ^{- p}}{p} + 1.

The lower bound uses the right inequality for

I_{N}

and adds

N^{- 1 - p}

; the upper uses the left inequality and adds one.

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