已知等差数列｛an｝的前n项和为Sn,且a1不等于0,求（n*an)/Sn的极限、（Sn+Sn+1）/(Sn+Sn-1)

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已知等差数列｛an｝的前n项和为Sn,且a1不等于0,求（n*an)/Sn的极限、（Sn+Sn+1）/(Sn+Sn-1)的极限

设：等差数列｛an｝的公差为 d ,通项为 an = a1+(n-1)d , 则：
sn = a1+a2+...+an = na1 +n(n-1)d/2
lim(n->∞) (n*an)/Sn
=lim(n->∞) [n*(a1+(n-1)d]/[na1 +n(n-1)d/2 ]
=lim(n->∞) [n*a1+n(n-1)d]/[na1 +n(n-1)d/2 ]
=lim(n->∞) [a1/(n-1) + d]/[a1/(n-1) + d/2 ]
= 2
lim(n->∞) (Sn+Sn+1）/(Sn+Sn-1)
=lim(n->∞) 1+ (an+an+1)/(Sn+Sn-1)
=lim(n->∞) 1+ {[a1 +nd] + [a1 +(n+1)d]}/{[na1 +n(n-1)d/2] + [(n-1)a1 +(n-1)(n-2)d/2]}
= 1+ 0
= 1

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