Prove that:
((lim(n->0)) (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)...(2n/(2n - 1))(2n/(2n + 1)) = pi/2
I know that I need to use
\(\displaystyle \int_0^(pi/2)\) (sinx)^(2n +1)dx = (2*4*....2n) / (3*5*...(2n + 1))
\(\displaystyle \int_0^(pi/2)\) (sinx)^(2n)dx = (pi/2) * ((1*3*5*...(2n - 1)) / (2*4*....2n))
How should I proceed? Thanks!
((lim(n->0)) (2/1)(2/3)(4/3)(4/5)(6/5)(6/7)...(2n/(2n - 1))(2n/(2n + 1)) = pi/2
I know that I need to use
\(\displaystyle \int_0^(pi/2)\) (sinx)^(2n +1)dx = (2*4*....2n) / (3*5*...(2n + 1))
\(\displaystyle \int_0^(pi/2)\) (sinx)^(2n)dx = (pi/2) * ((1*3*5*...(2n - 1)) / (2*4*....2n))
How should I proceed? Thanks!