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Question

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $ζ (2) = \sum_{k = 1}^{\infty} \frac{1}{k^{2}} = \frac{π^{2}}{6} .$

However, Euler was Euler and he gave other proofs.

I believe many of you know some nice proofs of this, can you please share it with us?

Accepted Answer

OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9 (EDIT: ...which is actually the proof that I read in Aigner & Ziegler).

When $0 < x < π / 2$ we have $0 < \sin x < x < \tan x$ and thus $\frac{1}{\tan^{2} x} < \frac{1}{x^{2}} < \frac{1}{\sin^{2} x} .$ Note that $1 / \tan^{2} x = 1 / \sin^{2} x - 1$ . Split the interval $(0, π / 2)$ into $2^{n}$ equal parts, and sum the inequality over the (inner) "gridpoints" $x_{k} = (π / 2) \cdot (k / 2^{n})$ : $\sum_{k = 1}^{2^{n} - 1} \frac{1}{\sin^{2} x_{k}} - \sum_{k = 1}^{2^{n} - 1} 1 < \sum_{k = 1}^{2^{n} - 1} \frac{1}{x_{k}^{2}} < \sum_{k = 1}^{2^{n} - 1} \frac{1}{\sin^{2} x_{k}} .$ Denoting the sum on the right-hand side by $S_{n}$ , we can write this as $S_{n} - (2^{n} - 1) < \sum_{k = 1}^{2^{n} - 1} {(\frac{2 \cdot 2^{n}}{π})}^{2} \frac{1}{k^{2}} < S_{n} .$

Although $S_{n}$ looks like a complicated sum, it can actually be computed fairly easily. To begin with, $\frac{1}{\sin^{2} x} + \frac{1}{\sin^{2} (\frac{π}{2} - x)} = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x \cdot \sin^{2} x} = \frac{4}{\sin^{2} 2 x} .$

Therefore, if we pair up the terms in the sum $S_{n}$ except the midpoint $π / 4$ (take the point $x_{k}$ in the left half of the interval $(0, π / 2)$ together with the point $π / 2 - x_{k}$ in the right half) we get 4 times a sum of the same form, but taking twice as big steps so that we only sum over every other gridpoint; that is, over those gridpoints that correspond to splitting the interval into $2^{n - 1}$ parts. And the midpoint $π / 4$ contributes with $1 / \sin^{2} (π / 4) = 2$ to the sum. In short, $S_{n} = 4 S_{n - 1} + 2 .$ Since $S_{1} = 2$ , the solution of this recurrence is $S_{n} = \frac{2 (4^{n} - 1)}{3} .$ (For example like this: the particular (constant) solution $(S_{p})_{n} = - 2 / 3$ plus the general solution to the homogeneous equation $(S_{h})_{n} = A \cdot 4^{n}$ , with the constant $A$ determined by the initial condition $S_{1} = (S_{p})_{1} + (S_{h})_{1} = 2$ .)

We now have $\frac{2 (4^{n} - 1)}{3} - (2^{n} - 1) \leq \frac{4^{n + 1}}{π^{2}} \sum_{k = 1}^{2^{n} - 1} \frac{1}{k^{2}} \leq \frac{2 (4^{n} - 1)}{3} .$ Multiply by $π^{2} / 4^{n + 1}$ and let $n \to \infty$ . This squeezes the partial sums between two sequences both tending to $π^{2} / 6$ . Voilà!

I might add that, as an alternative, one can evaluate the equivalent sum $\sum_{m = 0}^{\infty} (2 m + 1)^{- 2} = π^{2} / 8$ by summing only over the odd-numbered gridpoints. Then the midpoint $π / 4$ never enters the computation, and one gets an even simpler recurrence, of the form $T_{n} = 4 T_{n - 1}$ . — Oct 30, 2010 at 21:20
@Downvoter: Well, yes, at least from a modern perspective, since we define series using limits. I don't know if Euler thought about it that way. What's your point? — Nov 12, 2011 at 10:13
@Downvoter: it's hard to know whether you're really serious, but if so...Euler probably did more calculus-y things than any other mathematician in history (including Newton and Leibniz). — Mar 4, 2012 at 19:36
@AkivaWeinberger: Just saw this (sorry it's 3 years late), but I must have been, because I'm not sure what else I could've been thinking either... — Feb 19, 2017 at 22:20

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