数列极限问题
现在知道Sn=[(n+3)`n]/2
①求解1/S1+1/S2+```1/Sn的极限
答案是 11/9
②为什么1/Sn
人气：272 ℃ 时间：2020-03-27 05:52:11
解答
Sn=[(n+3)n]/21/Sn=2/[n(n+3)]=(2/3)[1/n -1/(n+3)]1/S1+1/S2+...+1/Sn=(2/3)[1/1-1/4+1/2-1/5+...+1/n-1/(n+3)]=(2/3)[(1/1+1/2+...+1/n)-(1/4+1/5+...+1/(n+3))]=(2/3)[1+1/2+1/3 -1/(n+1)-1/(n+2)-1/(n+3)]=11/9 ...感谢！ 还有第二小题1/Sn<2/[n(n+3)]1/S1<(2/3)(1/1-1/4)1/S2<(2/3)(1/2-1/5)…………1/Sn<(2/3)[1/n -1/(n+3)]累加1/S1+1/S2+...+1/Sn<(2/3)[1/1-1/4+1/2-1/5+...+1/n-1/(n+3)]1/S1+1/S2+...+1/Sn<11/9 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)](2/3)[1/(n+1)+1/(n+2)+1/(n+3)]>011/9-(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]<11/9-0=11/91/S1+1/S2+...+1/Sn<11/9 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]<11/9
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