由 yes 於 星期五 十一月 18, 2005 2:01 pm
左式=2^S
其中S=(1/2)+(2/2^2)+(3/2^3)+……+(n/2^n) -------(1)
(1/2)S=(1/2^2)+(2/2^3)+(3/2^4)+……+[(n-1)/2^n]+[n/2^(n+1)] -------(2)
(1)式-(2)式則(1/2)S=(1/2)+(1/2^2)+(1/2^3)+……+(1/2^n)-[n/2^(n+1)]
=(1/2)[1-(1/2)^n]/(1-1/2)- [n/2^(n+1)]=1-(1/2)^n- [n/2^(n+1)]
所以S=2-(1/2)^(n-1)- (n/2^n)<2
即左式<2^2=4故得證