前n项求和 an=arctan(1/（2n^2))

数学归纳法

S(1) = a(1) = arctan(1/2),
S(2) = a(1) + a(2) = arctan(1/2) + arctan(1/8),
tan[S(2)] = (1/2 + 1/8)/[1 - (1/2)(1/8)] = 10/15 = 2/3,
S(2) = arctan(2/3),
S(3) = S(2) + a(3) = arctan(2/3) + arctan[1/18]
tan[S(3)] = (2/3+1/18)/[1-(2/3)(1/18)] = 13*3/(26*2) = 3/4.
S(3) = arctan(3/4).

推测 S(n) = arctan[n/(n+1)]
=arctan [2n/(2n+2)]
=arctan {[(2n+1)-1]/[(2n+1)+1]}
=arctan {[tan arctan(2n+1) -tan π/4]/[1+ tan arctan(2n+1) • tan π/4]}
=arctan {tan [arctan(2n+1) - π/4]}
=arctan(2n+1) - π/4.

假设当n=k时，
Sn=arctan(2k+1)-π/4成立，则
tan S(k)=[tan arctan(2k+1) -tan π/4]/[1+tan arctan(2k+1) • tan π/4]
=[tan arctan(2k+1) - 1]/[tan arctan(2k+1) + 1]
=[(2k+1) - 1]/[(2k+1) + 1]
=2k/（2k+2）=k/（k+1）
则当n=k+1时，
tan S(k+1)=[tan S(k) + tan a（k+1）]/[1- tan S(k) • tan a（k+1）]
=[k/（k+1）+1/2（k+1）^2]/[1-(k/（k+1）)•(1/2（k+1）^2)]
=[2k（k+1）^2+（