Question

求解答第四题

求大神解

Answer 1

（1-1/2^2）(1-1/3^2)...(1-1/n^2)=(1-1/2)(1+1/2)(1-1/3)(1+1/3)...(1-1/n)(1+1/n)=1/2*3/2*2/3*4/3...[(n-1)/n]*[(n+1)/n]=1/2*[(n+1)/n]=(n+1)/2n
所以原式的极限为1/2

Answer 2

(lim省略)
=[[1+(1/2)][1+(1/3)]*……*[1+(1/n)]][[1-(1/2)][1-(1/3)]……[1-(1/n)]
=[(n+1)/2]*(1/n)
=(n+1)/2n