Question

This question may sound ridiculous at first sight, but let me please show you all how I arrived at the afore mentioned 'identity'.

Let us begin with (one of the many) equalities established by Euler:

$$ \displaystyle f(x) = \frac{sin(x)}{x} = \prod_{n=1}^{\infty} \Big(1-\frac{x^2}{n^2\pi^2}\Big) $$

as $(a^2-b^2)=(a+b)(a-b)$, we can also write: (EDIT: We can not write this...)

$$ \displaystyle f(x) = \prod_{n=1}^{\infty} \Big(1+\frac{x}{n\pi}\Big) \cdot \prod_{n=1}^{\infty} \Big(1-\frac{x}{n\pi}\Big) $$

We now we arrange the terms with $ (n = 1 \land n=-2)$, $ (n = -1 \land n=2$), $( n=3 \land -4)$ , $ (n=-3 \land n=4)$ , ..., $ (n = 2n \land n=-2n-1) $ and $(n=-2n \land n=2n+1)$ together . After doing so, we multiply the terms accordingly to the arrangement. If we write out the products, we get:

$$ f(x)=\big((1-x/2\pi + x/\pi -x^2/2\pi^2)(1+x/2\pi-x/\pi - x^2/2\pi^2)\big)... $$ $$ ...\big((1-\frac{x}{(2n)\pi} + \frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(n-1))^2\pi^2})(1+\frac{x}{2n\pi} -\frac{x}{(2n-1)\pi} -\frac{x^2}{(2n(2n-1))^2\pi^2)})\big) $$

Now we equate the $x^2$-term of this infinite product, using Newton's identities (notice that the'$x$'-terms are eliminated) to the $x^2$-term of the Taylor-expansion series of $\frac{sin(x)}{x}$ . So,

$$ -\frac{2}{\pi^2}\Big(\frac{1}{1\cdot2} + \frac{1}{3\cdot4} + \frac{1}{5\cdot6} + ... + \frac{1}{2n(2n-1)}\Big) = -\frac{1}{6} $$

Multiplying both sides by $-\pi^2$ and dividing by 2 yields

$$\sum_{n=1}^{\infty} \frac{1}{2n(2n-1)} = \pi^2/12 $$

That (infinite) sum 'also' equates $ln(2)$, however (According to the last section of this paper).

So we find $$ \frac{\pi^2}{12} = ln(2) $$ .

Of course we all know that this is not true (you can verify it by checking the first couple of digits). I'd like to know how much of this method, which I used to arive at this absurd conclusion, is true, where it goes wrong and how it can be improved to make it work in this and perhaps other cases (series).

Thanks in advance,

Max Muller

(note I: 'ln' means 'natural logarithm) (note II: with 'to make it work' means: 'to find the exact value of)

Answer 1

You cannot split (1-(x/n)^2) into (1 -x/n) (1 + x/n), since the products no longer converge.

Answer 2

Eisenstein defined elliptic functions by working with conditionally convergent series. In particular he studied how a series changes when you rearrange the terms in a specific way. You can find a lot about his work in this direction in Weil's beautiful book Elliptic Functions according to Eisenstein and Kronecker. An analogous question would be what happens to your product formula when you use a different way of pairing positive and negative indices. I do not know whether this has been studied before . . . A look into Weil's book will convince you (if you didn't know that already) that some functions are most interesting at those places where convergence fails.

Answer 3

It is a common trick, found in many elementary calculus texts: take a conditionally convergent series, and rearrange it to have any sum you like.