Binary Operations

Observe that $a*b=(a+1)(b+1)-1$

And $a*b*c = (a*b)*c = ((a+1)(b+1)-1)*c = \\ (((a+1)(b+1)-1)+1)(c+1)-1 = (a+1)(b+1)(c+1) -1$

Hence I am asserting that :

$\displaystyle { a }_{ 1 }*{ a }_{ 2 }*{ a }_{ 3 }*....*{ a }_{ n }=\prod _{ i=1 }^{ n }{ (1+{ a }_{ i }) } -1$

It can be easily proved by induction.

So what the question actually is asking is :

$\displaystyle S = \prod _{ i=1 }^{ \infty }{ (1+\dfrac { 1 }{ { i }^{ 2 } } ) } -1$

For calculating this consider the product expansion of $sin(x)$ :

$\displaystyle \dfrac{\sin { (x) }}{x} =\prod _{ n=1 }^{ \infty }{ (1-\dfrac { { x }^{ 2 } }{ { (n\pi ) }^{ 2 } } ) }$

Put $x= i \pi z$ to get :

$\displaystyle \dfrac{\sin { (i\pi z)}}{i\pi z} =\prod _{ n=1 }^{ \infty }{ (1+\dfrac { { z }^{ 2 } }{ { n }^{ 2 } } ) }$

Put $z=1$ to get :

$\displaystyle \dfrac{\sin { (i\pi )}}{i\pi } = \prod _{ n=1 }^{ \infty }{ (1+\dfrac { 1 }{ { n }^{ 2 } } ) }$

We will be finding $\sin(i\pi)$

Using euler's formula :

$e^{ix} = \cos(x)+i\sin(x)$

Put $-x$ instead of $x$

$e^{-ix} = \cos(x)-i\sin(x)$

Add subtract second from first to get :

$\displaystyle \sin(x) = \dfrac { { e }^{ ix }-{ e }^{ -ix } }{ 2i }$

Put $x=i\pi$ , to get :

$\displaystyle \sin(i\pi ) = \dfrac { { e }^{ -\pi }-{ e }^{ \pi } }{ 2i }$

$\Rightarrow \displaystyle \dfrac { \sin { i\pi } }{ i\pi } =\prod _{ n=1 }^{ \infty }{ (1+\dfrac { 1 }{ { n }^{ 2 } } ) } =\dfrac { { e }^{ \pi }-{ e }^{ -\pi } }{ 2\pi }$

Finally we have :

$\displaystyle S = \dfrac { { e }^{ \pi }-{ e }^{ -\pi } }{ 2\pi } -1$

Remarkable problem, and your maths skills regarding summations and integrals is ingenious,

However, i have found a bit of a cheap way to solve this problem, not half as awesome as yours

Simply expanding reveals that what we have to evaluate is

$(a+b+c+d...)\quad +\quad (ab+bc+cd+de....)+(abc+bcd+dce...)\quad .....\quad \infty$

here a,b,c,d are the reciprocal of squares

now the first sum is simply $\zeta (2)$

the second sum (sum over pairs) can be written as

$\\ \frac { \zeta (2)^{ 2 }-\zeta (4) }{ 2 }$

now consider the third sum, it can be seen with ease to be

$\\ \frac { \zeta (2)({ s }_{ 2 })-\zeta (4){ (s }_{ 1 })+\zeta (6) }{ 3 }$

where $s_{n}$ represnts the sum taken n tuples at a time

infact in general i found that it can be wriiten as

$s_{n}= \frac { 1 }{ n } \sum { \zeta ({ 2 }^{ r } } )({ s }_{ n-r })(-1)^{ r+1 }$

when we want to find the nth sum

though that is useless, as it is hard to sum over this,

However observing the first few terms revealed to me that they are simply

${ s }_{ r }=\frac { \pi ^{ 2(r-1) } }{(2 r-1)! }$

Thats it now it is obvious, simply consider the difference of two conjugate exponential series and divide by 'z', and proceed with ease to get answer,

Yes i agree that this method sucks