Sn-s(n-1)=n^2-(n-1)^2=2n-1=an(n≥2,n∈N)a1=1符合上式 an=2n-1,则a(n+1)=2n+1, 试写1/ana(n+1)=1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]=1/2[1/an-1/a(n+1)], 1/a(n-1)an=1/2[1/a(n-1)-1/an] 。
。。。。。。。。。。。。。。。。。。。。 1/(a2a3)=1/2(1/a2-1/a3) 1/(a1a2)=1/2(1/a1-1/a2) ∴Pn=1/2(1/a1-1/a2)+1/2(1/a2-1/a3)+。。。
+1/2[1/a(n-1)-1/an]+1/2[1/an-1/a(n+1)]=1/2[1/a1-1/a(n+1)] ∵a1=1,a(n+1)=2n+1∴Pn=1/2[1-1/(2n+1)] limPn=lim1/2[1-1/(2n+1)]=1/2(1-0)=1/2 n→+∞ n→+∞ 。
A1=S1=1，An=Sn-S(n-1)=2n-1, ∴ 1/AnA(n+1)=[1/(2n-1)-1/(2n+1)]/2 ∴ Pn=(1/2)[(1-1/3)+(1/3-1/5)+···+1/(2n-1)-1/(2n+1)] =(1/2)[1-1/(2n+1)], ∴ n→∞时,Pn→1/2.
an=Sn-S=2n-1 a=2n-3 1/a-1/an=(an-a)/ana=2/ana 所以1/ana=(1/2)(1/a-1/an) Pn=(1/2)[1/a1-1/a2+1/a2-1/a3+……+1/a-1/an] =(1/2)(1/a1-1/an) =(1/2)[1-1/(2n-1)] n趋向无穷时， Pn趋向1/2
Sn=n^2则a1=S1=1^2=1,an=Sn-S(n-1)=n^2-(n-1)^2=2n-1(n>=2)
故2/[ana(n+1)]
=2/[(2n-1)(2n+1)]
=[(2n+1)-(2n-1)]/[(2n-1)(2n+1)]
=1/(2n-1)-1/(2n+1)
因此1/[ana(n+1)]=(1/2)/(2n+1)-(1/2)/(2n-1)
所以Pn=(1/2){(1-1/2)+(1/2-1/3)+(1/4-1/4)+……+[1/(2n-1)-1/(2n+1)]}
=(1/2)[1-1/(2n+1)]
故n->∞limPn
=n->∞lim[1/2)-)1/2)/2n+1)]
==1/2-0
=1/2
答：(n+1)^3=n^3+3n^2+3n+1 --->(n+1)^3-n^3=3n^2+3n+1 n^3-(n-1)^3=3(n-1)^2+3(n-1)+1 (n...详情>>
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