Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 10

For

n \geq 6

and

n - 1 < x < n

, monotonicity gives

(x + 1)^{- 2} < n^{- 2} < x^{- 2}

. Integrating over each unit interval and summing proves the integral test tail bounds

\int_{5}^{\infty} \frac{d x}{( x + 1 ) ^{2}} = \frac{1}{6} < n = 6 \sum \infty \frac{1}{n ^{2}} < \int_{5}^{\infty} \frac{d x}{x ^{2}} = \frac{1}{5} .

The inequalities remain strict because the very first rectangle already has a strictly positive difference from either bounding integral.

ii

Add the supplied five-term partial sum to each side of the tail inequality. Since

1/6 = 600/3600

and

1/5 = 720/3600

\frac{5869}{3600} < n = 1 \sum \infty \frac{1}{n ^{2}} < \frac{5989}{3600} .

Numerically this is approximately

1.63028 < \sum n^{- 2} < 1.66361

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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