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Why is $\pi^2$ so close to~$10$?
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\\
Noam D. Elkies
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The ancient and still useful approximation $\pi \doteq \sqrt{10}$
may appear to be a mere coincidence.  But if we accept
Euler's theorem that
$$
\pi^2 = 6 \zeta(2) = 6 \sum_{n=1}^\infty \frac1{n^2}
$$
then we may easily deduce that $\pi^2<10$,
and estimate the error in the approximation.
Indeed we have
$$
\zeta(2) = 1 + \sum_{n=2}^\infty \frac1{n^2}
% < 1 + \sum_{n=2}^\infty \frac1{n^2-\frac14},
< 1 + \sum_{n=2}^\infty \frac4{4n^2-1},
$$
and the sum telescopes because
$$
% \frac1{n^2-\frac14}
% = \frac1{(n-\frac12)(n+\frac12)}
% = \frac1{n-\frac12} - \frac1{n+\frac12}.
\frac4{4n^2-1}
= \frac4{(2n-1)(2n+1)}
= \frac2{2n-1} - \frac2{2n+1}
= \frac2{2n-1} - \frac2{2(n+1)-1}.
$$
We conclude
$$
% \zeta(2) < 1 + \frac1{2-\frac12} = 1 + \frac23 = \frac53,
\zeta(2) < 1 + \Bigl(
   \bigl( \frac23-\frac25 \bigr) +
   \bigl( \frac25-\frac27 \bigr) +
   \bigl( \frac27-\frac29 \bigr) + \cdots
  \Bigr)
= 1 + \frac23 = \frac53,
$$
so $\pi^2 < 6 \cdot (5/3) = 10$ as claimed;
and the error is reasonably small because
$$
\frac53 - \zeta(2)
% = \sum_{n=2}^\infty \left( \frac1{n^2-\frac14} - \frac1{n^2} \right)
= \sum_{n=2}^\infty \left( \frac4{4n^2-1} - \frac1{n^2} \right)
= \sum_{n=2}^\infty \frac1{n^2(4n^2-1)},
$$
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 $\pi^2 = 9.8696044 \ldots$
is obtained by extracting a second term from the sum for $\zeta(2)$
% before bounding $1/n^2$ by $1/(n^2-\frac14)$: we find
before using $1/n^2 < 4/(4n^2-1)$: we find
$$
% \pi^2 = 6 \zeta(2) < 6 \left(1 + \frac14 + \frac1{3-\frac12}\right)
% = 6 \left(1 + \frac14 + \frac25\right) = 9.9 \, .
\pi^2 = 6 \zeta(2)
< 6 \left(1 + \frac14 + \frac25\right) = 9.9 \, .
$$
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