Why is π^{2} so close to 10?

Noam D. Elkies

The ancient and still useful approximation π ≐ \sqrt{10} may appear to be a mere coincidence. But if we accept Euler’s theorem that

π^{2} = 6ζ(2) = 6 ∑_{n=1}^{∞} \frac{1}{n^{2}}

then we may easily deduce that π^{2} < 10, and estimate the error in the approximation. Indeed we have

ζ(2) = 1 + ∑_{n=2}^{∞} \frac{1}{n^{2}} < 1 + ∑_{n=2}^{∞} \frac{4}{4n^{2} − 1},

and the sum telescopes because

\frac{4}{4n^{2} − 1} = \frac{4}{(2n − 1)(2n + 1)} = \frac{2}{2n − 1} − \frac{2}{2n + 1} = \frac{2}{2n − 1} − \frac{2}{2(n + 1) − 1}.

We conclude

ζ(2) < 1 + ((\frac{2}{3} − \frac{2}{5}) + (\frac{2}{5} − \frac{2}{7}) + (\frac{2}{7} − \frac{2}{9}) + ⋯) = 1 + \frac{2}{3} = \frac{5}{3},

so π^{2} < 6 · (5/3) = 10 as claimed; and the error is reasonably small because

\frac{5}{3} − ζ(2) = ∑_{n=2}^{∞} (\frac{4}{4n^{2} − 1} − \frac{1}{n^{2}} ) = ∑_{n=2}^{∞} \frac{1}{n^{2}(4n^{2} − 1)},

a sum whose first term is 1/60 and whose further terms are much smaller yet. An even better and still memorable approximation to the actual value π^{2} = 9.8696044 … is obtained by extracting a second term from the sum for ζ(2) before using 1/n^{2} < 4/(4n^{2} − 1): we find

π^{2} = 6ζ(2) < 6 (1 + \frac{1}{4} + \frac{2}{5}) = 9.9 .