Why is $ \frac{\pi^2}{12}=\ln(2)$ not true?

Question

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)

Answer 1

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.

Answer 2

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?

---

2. 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.)

Answer 3

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 4

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.