数列极限问题现在知道Sn=[(n+3)`n]/2①求解1/S1+1/S2+```1/Sn的极限答案是 11/9②为什么1
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
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
赞 爱尚在大洋彼岸 幼苗
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Sn=[(n+3)n]/2
1/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 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]
n->+∞,则n+1->+∞,n+2->+∞,n+3->+∞
1/(n+1)->0 1/(n+2)->0 1/(n+3)->0
(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]->0
11/9 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]->11/9
lim (1/S1+1/S2+...+1/Sn)=11/9
n->+∞
1/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 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]
n->+∞,则n+1->+∞,n+2->+∞,n+3->+∞
1/(n+1)->0 1/(n+2)->0 1/(n+3)->0
(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]->0
11/9 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]->11/9
lim (1/S1+1/S2+...+1/Sn)=11/9
n->+∞
1年前 追问
5
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)]>0 11/9-(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]<11/9-0=11/9 1/S1+1/S2+...+1/Sn<11/9 -(2/3)[1/(n+1)+1/(n+2)+1/(n+3)]<11/9
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