Using Leibniz's series pi/4 = 1 - 1/3 + 1/5 - 1/7 + .... prove that

pi/8 = 1/(1x3) + 1/(5x7) + 1/(9x11) + ... = 1/(2^2-1) + 1/(6^2-1) + 1/(10^2 - 1) + ...

So using the Leibniz's series we have pi/4 = (1 - 1/3) + (1/5 - 1/7) = (1/(2n-1) - 1/(2n+1) = 2/(2n-1)(2n+1)

Now I am stuck from here.... could you please help me complete this proof. Thanks.

pi/4 = (1 - 1/3) + (1/5 - 1/7) + ... + (1/(2n-1) - 1/(2n+1) + ...
= 2/(1x3) + 2/(5x7) + ... + 2((2n-1)x(2n+1)) + ...

pi/8 = pi/4 x 1/2 = 1/(1x3) + 1/(5x7) + ... + 1((2n-1)x(2n+1)) + ...
= /(2^2-1) + 1/(6^2-1) + 1/(10^2 - 1) + ...