Stolz定理
∞∗型Stolz定理
设bn+1−bnan+1−an=A∈Rcn=an−Abnbn+1−bncn+1−cn=bn+1−bn(an+1−Abn+1)−(an−Abn)=bn+1−bnan+1−anA→0故有bnan=bncn+A故不妨设A=0对任给ϵ>0,∃一个正整数N1,使得当n≤N1时,有−ϵ<bN1+1−bN1aN1+1−aN1<ϵ⋯−ϵ<bn−bn−1an−an−1<ϵ
−ϵ<bn−bN1an−aN1<ϵ−ϵ(bn−bN1)<an−aN1<ϵ(bn−bN1)所以存在一个正整数N≥N1,使得当n>N时,−2ϵ<bnan<2ϵ所以n→∞limbnan=0
A=+∞对任给M>0,∃N1,使得bn+1−bnan+1−an>M则当n>N1时,有bn−bN1an−aN1>Mbnan>bnaN1+M(1−bnbN1)>M=−所以,存在一个正整数N>N1,使得当n>N时bnan>2M所以n→∞limbnan=+∞
00型Stolz定理
{bn}严格减−ϵ<bn−bn+1an−an+1<ϵ则m>n>N时,有:−ϵ<bn−bman−am<ϵ固定n,令m→+∞后,−ϵ≤bnan≤ϵ