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

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

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

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

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

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:

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

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,

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

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 arrive 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)

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    $\begingroup$ Wouldn't it be cool if it were true, though! $\endgroup$ – Nate Eldredge Jun 9 '10 at 16:03
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    $\begingroup$ This forum is aimed at research-level mathematicians, but the topic of this question lies firmly in the domain of undergraduate mathematics. I think that "Ask Dr. Math" and "Art of Problem Solving" would be more appropriate venues for this question. $\endgroup$ – Ian Morris Jun 9 '10 at 17:41
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    $\begingroup$ @ Ian Morris: I guess you're right. I didn't expect my 'proof' to be torn apart that quickly. I thought that if the infinite product could be expressed by a product of two infinite products, I could produce documented mathematical thought on open problem(s) at research level. $\endgroup$ – Max Muller Jun 9 '10 at 17:56
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    $\begingroup$ Well, we all are safe once again. I'm a bit worried that young Max may eventually succeed in finding an inner contradiction in the building of Mathematics. Then, everybody go home, and this site will be terminated too :-( $\endgroup$ – Pietro Majer Jun 9 '10 at 21:09
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    $\begingroup$ @Ian Morris: Understanding when this type of calculation is valid is very definitely research-level. Most (valid) calculations of this type are actually short-hand for manipulating meromorphic functions, and it is often easier to guess the calculation first and then check it (this is why Euler was such a great mathematician). One place where these calculations turn up in spades is in analytic number theory, at least if the two weeks of the course I took form R. Borcherds is any indication. Another place is quantum field theory, where there are outstanding problems to define certain integrals. $\endgroup$ – Theo Johnson-Freyd Jun 12 '10 at 16:38
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You cannot split $$\left(1-\left(\frac{x}{n}\right)^2\right)\tag{1}$$ into $$\left(1 -\frac{x}{n}\right) \left(1 + \frac{x}{n}\right)\tag{2}$$ since the products no longer converge.

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  • $\begingroup$ Could you please elaborate on that? Why is it necessary for these prodcucts to converge? $\endgroup$ – Max Muller Jun 9 '10 at 16:21
  • $\begingroup$ See en.wikipedia.org/wiki/Infinite_product and en.wikipedia.org/wiki/Conditional_convergence $\endgroup$ – M.G. Jun 9 '10 at 16:31
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    $\begingroup$ The statements: - The product of (1 + a_n) converges - The sum of a_n converges are equivalent. So we know that prod (1 + a_n), where a_2n = (1 + x/n) and a_2n = (1 - x/n) does not converge unconditionally. So by reordering it can achieve any positive value by the same argument as for sums. Just rewrite product (1 + a_n) = exp(sum log(1 + a_n)) to show this. $\endgroup$ – Helge Jun 9 '10 at 16:50
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    $\begingroup$ Yes, you treated divergent infinite products as convergent. You get the same sort of problems as you do with treating divergent series as convergent, like $$0 = (1 - 1) + (1 - 1) +\cdots=1-(1-1)+(1-1)\cdots =1.$$ $\endgroup$ – Robin Chapman Jun 9 '10 at 17:40
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    $\begingroup$ (@Qiaochu: My apologies for misspelling your name.) $\endgroup$ – Harald Hanche-Olsen Jun 9 '10 at 23:54
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Maybe I'm too late to be of much use to the original question-asker, but I was surprised to see that all of the previous answers seem to not quite address the real point in this question.


  1. While it is important to be aware of the dangers of rearranging conditionally convergent series, it not true that any rearrangement is invalid, in terms of changing the value of the sum.

Namely, any finite rearrangement of terms will obviously leave the sum unchanged. So will any collection of disjoint finite rearrangements. For example, by a standard Taylor series argument the following sum converges conditionally to $\ln (2)$: $$ H_{\pm} = \sum_{m\geq 1}(-1)^{m+1}\frac1{m} = 1 - \frac12 + \frac13 - \frac14 + \cdots = \ln(2).$$ (Note that by grouping terms, $ H_{\pm} = \sum_{n\geq 1} \left( \frac{1}{2n-1} - \frac{1}{2n}\right)= \sum_{n\geq1} \frac1{2n(2n-1)}$.) The following ''rearranged'' sum also converges to the same value: $$ H_{\pm}^* = \sum_{n\geq 1} \left( - \frac{1}{2n} + \frac{1}{2n-1}\right) = -\frac12 + 1 - \frac14 + \frac13 - \cdots .$$ The partial sums of $H_{\pm}$ and $H_{\pm}^*$ share a subsequence in common, the partial sums indexed by even numbers, so the two series must both converge to the same value.

This is essentially the same type of rearrangement that Max is considering in the question. The product of $$ \textstyle\left(1 + \frac1{2n-1}\frac{x}{\pi}\right) \left(1 -\frac1{2n-1} \frac{x}{\pi} \right) \qquad\text{and}\qquad \left(1 + \frac1{2n}\frac{x}{\pi}\right) \left(1 - \frac1{2n} \frac{x}{\pi} \right) $$ can be rearranged as the product $$ \textstyle\left(1 + \frac1{2n-1}\frac{x}{\pi}\right) \left(1 -\frac1{2n} \frac{x}{\pi} \right) \qquad\text{and}\qquad \left(1 - \frac1{2n-1}\frac{x}{\pi}\right) \left(1 +\frac1{2n} \frac{x}{\pi} \right) .$$ This does not change the value of the (conditionally convergent) infinite product for $\frac{\sin x}{x}$.

So if the error in this ''proof'' of $\pi^2/6 = \ln(2)$ is not in rearranging terms, where is the actual mistake?


  1. The mistake is in leaving out a term when "foiling" the product of two polynomials. (Or, in a misapplication of Newton's identities.)

The valid infinite product expression $$ \frac{\sin x}{x} = \prod_{n\geq 1} \textstyle\left(1 + \frac1{2n-1} \frac x{\pi}\right)\left(1 -\frac1{2n} \frac{x}{\pi}\right) \left(1 - \frac1{2n-1}\frac{x}{\pi}\right)\left(1 +\frac1{2n}\frac{x}{\pi} \right) $$ $$\qquad\qquad \qquad\quad= \prod_{n\geq 1} \textstyle\left(1 + \frac1{2n(2n-1)} \frac{x}{\pi} - \frac1{2n(2n-1)}\frac{x^2}{\pi^2} \right) \left(1 - \frac1{2n(2n-1)} \frac{x}{\pi} - \frac1{2n(2n-1)}\frac{x^2}{\pi^2} \right) $$ simplifies to $$\frac{\sin x}{x} = \prod_{n\geq 1} \textstyle\left(1 - \frac{2}{2n(2n-1)}\frac{x^2}{\pi^2} - \frac{1}{(2n)^2(2n-1)^2} \frac{x^2}{\pi^2} + O(x^4)\right) .$$

The coefficient of $x^2$ in this product is the series $$ -\frac1{\pi^2} \sum_{n\geq 1}\left( \frac{2}{2n(2n-1)} + \frac{1}{(2n)^2(2n-1)^2} \right)$$ which does, in fact, converge to $ -\frac1{\pi^2}\zeta(2) = -\frac1 6$. This can be checked through some algebra, or by asking WolframAlpha.

(This explains why $ \sum_{n\geq 1} \frac{1}{(2n)^2(2n-1)^2} = \frac{\pi^2}{6}-2\ln(2),$ which I would not have known how to evaluate otherwise.)

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    $\begingroup$ As far as I can tell, you are correct and the other answers are irrelevant. This sort of regrouping is equivalent to writing $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} + \frac{1} {(2n)^2}$ and is completely harmless, if one doesn't make an algebra error. $\endgroup$ – Will Sawin Aug 23 '19 at 20:17
  • $\begingroup$ mathworld.wolfram.com/LevyConstant.html $\endgroup$ – VS. Feb 13 at 17:00
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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.

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  • $\begingroup$ Thank you for the reference! As for the different pairing methods, that was what I was thinking about as well... An infinite amount of different (infinite) sum series would all converge to the same value! By the way, do you think there's a way to describe my double product formula in a closed form? I know it diverges, so perhaps it could be an exponential function? Like sin(x)^x? $\endgroup$ – Max Muller Jun 10 '10 at 13:31
  • $\begingroup$ P.S. I'm not sure why (some) functions are most interesting at those places where convergence fails... could you please explain? I am very interested in transforming divergent to convergent series, though, to make the double product equal the single product (and sin(x)/x). Do you think this is possible, one way or another? Or do you know any texts/papers on this subject matter? $\endgroup$ – Max Muller Jun 10 '10 at 13:44
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    $\begingroup$ values for divergent series ... see Hardy's book, DIVERGENT SERIES $\endgroup$ – Gerald Edgar Jun 10 '10 at 15:17
  • $\begingroup$ I was of course thinking of zeta functions, L-series and theta functions. For methods of assigning finite values to interesting divergent sums (as well as for plenty of other reasons as well), I strongly advise you to read <em>Euler through time: a new look at old themes</em> by Varadarajan. $\endgroup$ – Franz Lemmermeyer Jun 10 '10 at 17:52
  • $\begingroup$ I thank both of you for the references! I think I'll have some good reading for the summer holidays... $\endgroup$ – Max Muller Jun 10 '10 at 18:30
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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.

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  • $\begingroup$ Ok, thanks mister Edgar. Please look at the comments I posted to Franz Lemmermeyer's answer, I'd like to know if you have some answers on the questions I posed there as well! $\endgroup$ – Max Muller Jun 10 '10 at 13:45
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    $\begingroup$ You probably already know this Edgar, but this is usually called the Riemann rearrangement theorem, or the Riemann series theorem $\endgroup$ – Daniel Barter Jun 13 '10 at 3:55

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