已知等差数列{an}的首项a1及公差d都是整数，前n项和为Sn，若a1>1,a4>3，S3≤9，设bn=1/nan，则使b1+b2+...
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a1+a2+a3 <= 9
1 3 5 is not an option as a1 > 1
so 2 3 4 is the only option for a1 a2 a3, so an = n+1
bn = 1/[n(n+1)] = 1/n - 1/(n+1)
b1+b2+...+bn = 1 - 1/(n+1)
if n = 99, b1+b2+...+bn = 1-1/100 = 99/100
so n = 98
1 3 5 is not an option as a1 > 1
so 2 3 4 is the only option for a1 a2 a3, so an = n+1
bn = 1/[n(n+1)] = 1/n - 1/(n+1)
b1+b2+...+bn = 1 - 1/(n+1)
if n = 99, b1+b2+...+bn = 1-1/100 = 99/100
so n = 98
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