求lim(Sn+Sn+1)/(Sn+Sn-1),an为等差数列,a1不为零

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设sn=a1+(n-1)d,则sn=a1*n+n(n-1)d/2
代入
lim(Sn+Sn+1)/(Sn+Sn-1)
=[na1+n(n-1)d/2+（n+1）*a1+n^2d]/[na1+n(n-1)d/2+（n-1）a1+(n-2)(n-1）d/2]
上下同时除以n^2
得
=lim[1/na1+(1-1/n)d/2+（1/n+1/n^2）*a1+d]/[1/na1+(1-1/n)d/2+（1/n-1/n^2）a1+(1-3/n+2/n^2）d/2]
显然,当n趋向无穷大时,1/n,1/n^2均为0
则可化简为
(d/2+d/2)/(d/2+d/2)
=1

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