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The problem of determining the value of $\zeta (2)$ also known as t...
The term Telescoping sum applies to an expression of the form $$...
The upper bound idea is quite clever, but the lower bound (which is...
There's also a geometric illustration of why $\pi\approx \sqrt{10}$...
If you think about it, this value is also very close to the gravita...
Why is π
2
so close to 10?
Noam D. Elkies
The ancient and still useful approximation π
.
=
10 may appear to be a mere
coincidence. But if we accept Euler’s theorem that
π
2
= 6ζ(2) = 6
X
n=1
1
n
2
then we may easily deduce that π
2
< 10, and estimate the error in the approx-
imation. Indeed we have
ζ(2) = 1 +
X
n=2
1
n
2
< 1 +
X
n=2
4
4n
2
1
,
and the sum telescopes because
4
4n
2
1
=
4
(2n 1)(2n + 1)
=
2
2n 1
2
2n + 1
=
2
2n 1
2
2(n + 1) 1
.
We conclude
ζ(2) < 1 +
2
3
2
5
+
2
5
2
7
+
2
7
2
9
+ · · ·
= 1 +
2
3
=
5
3
,
so π
2
< 6 · (5/3) = 10 as claimed; and the error is reasonably small because
5
3
ζ(2) =
X
n=2
4
4n
2
1
1
n
2
=
X
n=2
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 +
1
4
+
2
5
= 9.9 .

Discussion

The term Telescoping sum applies to an expression of the form $$ \sum_{k=0}^{n} (a(k+1)−a(k)) $$ which ends up being equal to $a(n+1)-a(0)$. It's easy to see since $$ \sum_{k=0}^{n} (a(k+1)−a(k))= \sum_{k=1}^{n+1} a(k)−\sum_{n=0}^{n}a(k) = a(n+1) - a(0) $$ For infinite series the telescoping series concept exists as long as $\lim_{n \rightarrow \infty} a(n)$ exists. In this specific case $\lim_{n \rightarrow \infty} a(n) = \lim_{n \rightarrow \infty} \frac{2}{2n-1}=0 $ The upper bound idea is quite clever, but the lower bound (which is needed to show how far from 10 we really are) is even more important -- and can be made explicit in a similar way. Just note: $\sum_{n=2}^{\infty} \frac{1}{n^2(4n^2 - 1)} < \sum_{n=2}^{\infty} \frac{1}{60/16 n^4} < \int_1^{\infty} \frac{16}{60} t^{-4} dt = \frac{16}{60} \cdot \frac{1}{3} = \frac{4}{45}.$ Therefore $\frac{5}{3} - \zeta(2) < \frac{4}{45}$, or $\zeta(2) > \frac{5}{3} - \frac{4}{45} = \frac{71}{45}$, and multiplying by 6 we get: $\pi^2 > \frac{426}{45} = 9.466$. The problem of determining the value of $\zeta (2)$ also known as the Basel problem was first posed by [Pietro Mengoli](https://en.wikipedia.org/wiki/Pietro_Mengoli) in 1644. It consisted of determining the value of the following series $$ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} $$ A lot of famous mathematicians tried to attack this problem but was Euler at 28 in 1735 who proved for the first time that $\zeta(2) = \frac{\pi^2}{6}$ (the problem is named after Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem). To prove it Euler used the Maclaurin series of the sine function $$ \sin (\pi x) = \pi x - \frac{(\pi x)^3}{3!}+\frac{(\pi x)^5}{5!}- \ldots =: p(x) $$ Since the roots of $sin(\pi x)$ are the integers $\mathbb{Z}$ and for finite polynomials p(x) we know that we can write the function as a product of linear factors of the form $(1 −\frac{x}{a})$, where $p(a) = 0$. Euler conjectured that the same trick would work here for $\sin(πx)$. Assuming, for the moment, that this is correct, we have $$ p(x) = \pi x\left(1-\frac{x}{1}\right)\left(1+\frac{x}{1}\right)\left(1-\frac{x}{2}\right)\left(1+\frac{x}{2}\right)\ldots \\ = \pi x\left(1-\frac{x^2}{1}\right)\left(1-\frac{x^2}{4}\right)\left(1-\frac{x^2}{9}\right) \ldots \\ = \pi x \prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right) $$ If we examine the coefficient of $x^3$ in the formula, we have $$ -\pi\left(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\ldots\right) = -\pi \sum_{n=1}^{\infty}\frac{1}{n^2} $$ Now we just have to compare this with the Maclaurin series to get $$ -\frac{\pi^3}{6} = - \pi \sum_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2} $$ It is important to note that we can perform the root factorization on the sine function because of the Weierstrass factorization theorem which states that we can do it for any entire function over $\mathbb{C}$. We also annotated on Fermat's Library [a paper by Tom Apostol](http://fermatslibrary.com/s/a-proof-that-euler-missed) that gives an alternative proof to calculating $\zeta(2)$ based on solving a double integral which avoids Euler's expansions of the sine function. If you think about it, this value is also very close to the gravitational acceleration near Earth's surface :) In fact, in 1668 the English cleric John Wilkins proposed a decimal-based unit of length, on a pendulum with a 2 second period. ![](http://www.webassign.net/question_assets/asucolphysmechl1/lab_7/images/figure7-1-alt.png) For a pendulum with a small angle at the surface of the Earth the period of oscillation is $$ T= 2 \pi \sqrt{\frac{L}{g}} $$ If we set the period T to 2 seconds we get $$ 2=2\pi\sqrt{\frac{L}{g}} \equiv L = \frac{g}{\pi^2} $$ Now if we call L, 1 meter we have $\pi^2 \approx g$ and that's why the values are related! There's also a geometric illustration of why $\pi\approx \sqrt{10}$. Assuming we have a section of a circle with angle $\theta = \frac{\pi}{10}$ and radius 1. ![](https://i.imgur.com/OJZaGAP.png) Using the quintuple-angle formula we have $$ 0 = \cos (5\theta) = 5 \cos (\theta) - 20 \cos^3(\theta) + 16 \cos^5 (\theta) \\ = 5(1-x)-20(1-x)^3+16(1-x)^5 \\ = 1-25x+100x^2+O(x^3) $$ Since $x << 1$, we can neglect the higher terms and solve the quadratic for $x \approx \frac{1}{20}$. Using the Pythagorean theorem we have $\theta \approx \sqrt{h^2+x^2} = \sqrt{1-(1-x)^2+x^2} = \sqrt{2x}$ and so $\pi \approx \sqrt{10}$.