(n^3-1)/(n^3+1) 从n=2累乘到正无穷的极限怎么求？

(n^3-1)/(n^3+1) 从n=2累乘到正无穷的极限怎么求？

(n^3-1)/(n^3+1) = (n-1)(n^2+n+1) / [(n+1)(n^2-n+1)]
= (n-1)n(n^2+n+1) / [n(n+1)(n^2-n+1)]
= (n-1)n/(n^2-n+1) / [n(n+1)/(n^2+n+1)]
= (n-1)n/(n(n-1)+1) / [n(n+1)/((n+1)n+1)]
= a[n-1] / a[n]
其中a[n] = n(n+1)/((n+1)n+1)
所以a1/a2 * a2/a3 * ··· * a[n-1]/a[n] = a1/a[n] = 2/3 / a[n]
注意到a[n]→1
所以这个极限为2/3

lim(-1+3^n)/2^(n+1) =lim[(-1/3^n)+1/[(2/3)^n*2] 当n趋向正无穷时 3^n趋向正无穷 所以-1/3^n趋向负无穷小 而(2/3)^n趋向正无穷小 所以 分子是1+负无穷小 分母是正无穷小 所以 lim(-1+3^n)/2^(n+1)[n趋向正无穷] 趋向正无穷