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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

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.

Figure 1.
Unit rectangles between two shifted reciprocal-square curves
. Original comparison for the first five intervals of the infinite tail. Each rectangle has height $1/ n^{2}$ on $[n - 1, n]$ , above $(x + 1)^{- 2}$ and below $x^{- 2}$ . Details.

ii

Solution

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

.

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