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)

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)

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)

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)

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 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)

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)