lim(an+1/an)=p,p

lim(an+1/an)=p,p 如果级数收敛,则Lim (an+1/an)=p 数列{an}满足an=p/(pn+1-p)(0 如何求这两种数列的通项 an-1 - an=p*an-1*an an-1^p=an lim|(an+1)/an|=q 已知数列{an}满足a0=1,an=p|an-1|-1(n∈N,p为常数,0 lim n趋于无穷 ,an+1/an=q.求lim an=? lim(an+1-an)=d,求证lim(an/n)=d 极限的简单运算lim an=3, lim bn=1/3, 求lim ((an-3bn)/2an) lim(2an+4bn)=1 lim(3an-bn)=2 求lim(an+bn) lim(2an+4bn)=1 lim(3an-bn)=2 求lim(an+bn) 已知数列{An}对于任意p,q属于N*,有Ap+Aq=A(p+q)+1/p(p+q),若a1=1,则An= {An}{Bn},若lim(3An+Bn)=8,lim(6An-Bn)=1求lim(4An-Bn)与lim(AnBn)的值 {an}{bn}中,a1=2,b1=4,an,bn,an+1成A,P,bn,an+1,bn+1成G,P 求an,bn.证明(1/a1+b1)+(1/a2+b2)+...+1/an+bn 已知数列{an}满足a1=1,a2=a(a≠0)an+2=p×（an+1）²/an（其他p为非零常数n∈N*）判断数列{an+1/an}时不是等比数列 f(x)=负的根号下4+x的平方,p(an,-1/an),求an (x)=负的根号下4+（x的平方）分之一,p(an,-1/an),求an 数列{an}的前 n 项和为 Sn,p 为常数,且满足 a1=2,Sn=4an+Sn-1-pan-1(n≥2), 又已数列{an}的前 n 项和为 Sn,p 为常数,且满足a1=2,Sn=4an+Sn-1-pan-1(n≥2), 又已知 lim S n = 6 ,则 p= ?（答案为2,求过程~谢谢!）