数列{an},an=1/[n*2^(n-1)].前N项和为Sn,求证Sn

题目

数列{an},an=1/[n*2^(n-1)].前N项和为Sn,求证Sn

答案

证明：
∵当n>1时,(n-2)2^(n-1)≥0
∴n2^(n-1)-2^n≥0
n2^(n-1)≥2^n
即：1/[n2^(n-1)]≤1/2^n
∵数列{a[n]},a[n]=1/[n2^(n-1)],前n项和为S[n]
∴S[n]
=1+1/(2*2^1)+...+1/[n2^(n-1)]
≤1+1/2^2+...+1/2^n
=1+(1/4)[1-1/2^(n-1)]/(1-1/2)
=1+(1/2)[1-1/2^(n-1)]
＜1+1/2
=3/2