证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)有人说有五种方法,我已经得到三种,请大神再提供

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证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)有人说有五种方法,我已经得到三种,请大神再提供证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)有人说有五种方法,

证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)有人说有五种方法,我已经得到三种,请大神再提供
证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)
有人说有五种方法,我已经得到三种,请大神再提供

证明1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)有人说有五种方法,我已经得到三种,请大神再提供
consider
for n>=3
1/(2^n-1) < (2/3) [1/2^(n-1)]
an = 1/[2^(n+1) -1]
a1=1/3
1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)
=Sn
=a1+a2+..+an
= 1/3 + 1/(2^3-1)-1/(2^4-1)+...+1/(2^n-1)

1/(2^2-1)+1/(2^3-1)+……+1/(2^n-1)<1/2^2+1/2^3+……+1/2^n
=(1/2^2)*(1-(1/2)^n)/(1-1/2)<1/2<2/3