Question

微积分题目？

Answer 1

原式 = (1/2)∑<n=1,∞>[1/(2n-1) - 1/(2n+1)]
= (1/2)[1-1/3 + 1/3-1/5 + 1/5-1/7 + 1/7-1/9 +......]
= 1/2 , 当然原级数收敛。

Answer 2

∑(n:1->∞) 1/[(2n-1)(2n+1)]
=(1/2)∑(n:1->∞) [1/(2n-1)- 1/(2n+1)]
=(1/2) [ 1 - lim(n->∞) 1/(2n+1) ]
=(1/2)(1-0)
=1/2