数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn

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数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn
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数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn
数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn<3/4急用!

数列An为等差数列,其前n项和为Sn 且a3=7,a5=11.求证1/s1+1/s2+.+1/Sn
设通项公式为an=a1+(n-1)d
11-7=a5-a3=(5-3)d
2d=4,
d=2
a3=a1+2d=a1+4=7
a1=3
Sn=n(n+2)
1/S1+1/S2+1/S3+……+1/Sn
=1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+……+1/[n(n+2)]
=(1/2)[(1/1-1/3)+(1/2-1/4)+(1/3-1/5）+……+1/n-1/(n+2)]
=(1/2)[1/1+1/2-1/(n+1)-1/(n+2)]


a5-a3=2d
a5=11 a3=7代入
2d=11-7=4 d=2
a1=a3-2d=7-4=3
数列{an}是以3为首项，2为公差的等差数列。
Sn=3n+2n(n-1)/2=n²+2n=n(n+2)
1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
1/S1+1/S2+...+1/Sn

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a5-a3=2d
a5=11 a3=7代入
2d=11-7=4 d=2
a1=a3-2d=7-4=3
数列{an}是以3为首项，2为公差的等差数列。
Sn=3n+2n(n-1)/2=n²+2n=n(n+2)
1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
1/S1+1/S2+...+1/Sn
=(1/2)[1-1/3+1/2-1/4+...+1/n-1/(n+2)]
=(1/2)[(1+1/2+1/3+...+1/n)-(1/3+1/4+...+1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(1/2)[3/2-1/(n+1)-1/(n+2)]
=3/4-(1/2)[1/(n+1)+1/(n+2)]
1/(n+1)>0 1/(n+2)>0
1/(n+1)+1/(n+2)>0
1/S1+1/S2+...+1/Sn<3/4-0
1/S1+1/S2+...+1/Sn<3/4
不等式成立。

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a3=7,a5=11
a1=3 d=2
an=2n+1
sn=n(n+2)
1/sn=1/2(1/n-1/(n+2)
1/s1+1/s2+.....+1/Sn=1/2[1-1/3+1/2-1/4+1/3+……-1/n+1/(n+1)-1/(n+2)
=1/2[3/2-1/n-1/(n+2)]<3/4

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