1-1/3+1/5-1/7+1/9-...=pi/4, according to Leibniz series for arctan expansion, also known as the Gregory's formula.

here is my modification (for the sum above) of euler's proof for zeta(2). see if this makes sense.
sin(x)-1=0
gives solutions for x=pi/2+2pik, or x=pi/2, 5pi/2, 9pi/2,... and x=-3pi/2, -7pi/2,-11pi/2

sin(x)-1=-1+x-x^3/3!+x^5/5!-... by the taylor series

if we write sin(x)-1=0 as -(1-x/r1)(1-x/r2)...=0

we get sin(x)-1=-(1-2x/pi)(1+2x/(3pi))(1-2x/(5pi))(1+2x/(7pi))...=-1+x-x^3/3!+x^5/5!-...

so the coefficient in front of x should be the same:

or -2/pi+2/(3pi)-2/(5pi)+2/(7pi)-...=-1
1-1/3+1/5-1/7=pi/2 ??? similar, but twice as large as pi/4
whats wrong with my logic?