Permalink Submitted by Peter L. Griffiths on June 16, 2016

The proof of the sine Basel conjecture (PI)^2/2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .... depends on the Newtonian Infinite Series formulae which are
the ABC summation 1 + Ax + Bx^2 + Cx^3 .... = (1+ax)(1+ bx)(1 +cx)....
the ABC Alternating 1 -Ax + Bx^2 - Cx^3 .... = (1- ax)(1- bx) (1- cx) ...
the A summation Ax = ax + bx + cx..... which seems to be a special case of the sine Basel conjecture (PI)^2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .....
with both sides multiplied by (u/PI)^2 and thus becomes u^2/6 = (u/PI)^2 + (u/2PI)^2 + (u/3PI)^2 .... clearly a version of the A summation
Au^2 = au^2 + bu^2 + cu^2 ... with A = 1/6, a =(1/PI)^2, b = (1/2PI)^2, c = (1/3PI)^2.
The alternating sine series is (sinu)/u = 1 - u^2/3! + u^4/5! - u^6/7!.....clearly a special case of the ABC Alternating, whose product series can now be
evaluated by substituting values for letters.

## Euler's Basel Conjecture

The proof of the sine Basel conjecture (PI)^2/2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .... depends on the Newtonian Infinite Series formulae which are

the ABC summation 1 + Ax + Bx^2 + Cx^3 .... = (1+ax)(1+ bx)(1 +cx)....

the ABC Alternating 1 -Ax + Bx^2 - Cx^3 .... = (1- ax)(1- bx) (1- cx) ...

the A summation Ax = ax + bx + cx..... which seems to be a special case of the sine Basel conjecture (PI)^2 = 1 + 1/2^2 + 1/3^2 + 1/4^2 .....

with both sides multiplied by (u/PI)^2 and thus becomes u^2/6 = (u/PI)^2 + (u/2PI)^2 + (u/3PI)^2 .... clearly a version of the A summation

Au^2 = au^2 + bu^2 + cu^2 ... with A = 1/6, a =(1/PI)^2, b = (1/2PI)^2, c = (1/3PI)^2.

The alternating sine series is (sinu)/u = 1 - u^2/3! + u^4/5! - u^6/7!.....clearly a special case of the ABC Alternating, whose product series can now be

evaluated by substituting values for letters.