Question

求和。。。谢谢！！

Answer 1

http://mathforum.org/library/drmath/view/60401.html
看明白了吗？
因为：tan(A-B)=(tanA-tanB)/(1+tanA*tanB)
设 tanA = n+1 tanB= n
tan(A-B) = (n+1-n)/(1+n(n+1)) = 1/n^2+n+1
arctan(1/n^2+n+1) = A-B = arctan(n+1) - arctan(n)
设U(n)=arctan(1/n^2+n+1)
U(1) = arctan(2) - arctan(1)
U(2) = arctan(3) - arctan(2)
.....
U(n) = arctan(n+1) - arctan(n)
.....

U(1)+U(2)+.......+U(n)+....U(∞)=arctan(∞+1) - arctan(1)
=arctan(∞)-arctan(1) =π/2 -π/4 =π/4