Question

求解析大学微积分

Answer 1

Answer 2

an = ln(n+1)/(n+1)
lim(n->∞) a(n+1)/an
=lim(n->∞) [ln(n+2)/(n+2)]/[ln(n+1)/(n+1)]
=lim(n->∞) [ln(n+2)/ln(n+1)] . lim(n->∞) (n+1)/(n+2)
=(1).(1)
=1

追问

为什么lim(n->∞) [ln(n+2)/ln(n+1)]等于1

为什么lim(n->∞) [ln(n+2)/ln(n+1)]等于1

追答

lim(x->∞) ln(2+x)/ln(1+x) (∞/∞ 分子分母分别求导)

=lim(x->∞) [1/(2+x)]/[1/(1+x)]
=lim(x->∞) (x+1)/(x+2)
=1
=>lim(n->∞) [ln(n+2)/ln(n+1)]=1