 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1/10/b/solution

Apply the integral bounds for a finite harmonic sum to the omitted terms from thirteen through one hundred:

ln \frac{101}{13} < \sum_{n = 13}^{100} \frac{1}{n} < ln \frac{100}{12} .

Adding the given

H_{12} = 86021/27720

gives

5.15338 \dots < H_{100} < 5.22348 \dots

. Both bounds lie between

5.15

and

5.25

, so every value they permit rounds to the same two significant figures:

H_{100} = n = 1 \sum 100 \frac{1}{n} ≃ 5.2 to two significant figures .

The rounding follows from the bounds, without computing all the intervening terms.

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