1/(1+t^2) = 1 - t^2 + t^4 - ... + (-1)^(n-1) * t^(2n - 2) + (-1)^n * t^(2n) / (1+t^2) for any integer n >= 1
(b) For any x ∈ (-1, 1), prove that
arctan x = [x - x^3 / 3 + x^5 / 5 - ... + (-1)^(n-1) x^(2n-1) / (2n - 1)] + (-1)^n ∫(0 to x) t^(2n) / (1+t^2) dt for any positive integer n.
Hence, show that
| arctan x - [x - x^3 / 3 + x^5 / 5 - ... + (-1)^(n-1) x^(2n-1) / (2n - 1)] | <= [|x|^(2n+1)] / (2n+1)
(c) Prove that
arctan x = lim_n->∞ [x - x^3 / 3 + x^5 / 5 - ... + (-1)^(n-1) x^(2n-1) / (2n-1)] for -1 <= x <= 1. Hence evaluate lim_n->∞ [1 - 1/3 + 1/5 - ... + (-1)^(n-1) 1/(2n-1)]
更新1:
For (a) part, I think there is a simplier method. Using the fact (1 + k)(1 - k + k^2 - ... + (-1)^(n-1) k^(n-1) = 1 + (-1)^(n-1) k^n 1 - k + k^2 - ... + (-1)^(n-1) k^(n-1) = 1/(1+k) + [(-1)^(n-1) k^n]/(1+k) 1/(1+k) = 1 - k + k^2 - ... + (-1)^(n-1) k^(n-1) + [(-1)^n k^n]/(1+k)
更新2:
then put k = t^2, the proof follows.
