Question

求数列{2n-1/2n}的前n项和Sn

求数列{2n-1/2n}的前n项和Sn

Answer 1

答案是n减去一半的n的调和数，当n足够大的时候答案约为n-(ln[n]-r),其中r为欧拉常数，r=0.5772.....

Answer 2

设{(2n-1)/2^n}的前n项和为Sn.
Sn=1/2+3/2^2+5/2^3+…+(2n-1)/2^n (1)
(1/2)*(1)得：(1/2)Sn=1/2^2+3/2^3+5/2^4+…+(2n-1)/2^(n+1) (2)
(1)-(2)得：(1/2)Sn=1/2+1/2^2+1/2^3+…+1/2^n-(2n-1)/2^(n+1)
=(1/2)(1-1/2^n)/(1-1/2)-(2n-1)/2^(n+1)
=1-2/2^(n+1)-(2n-1)/2^(n+1)
=1-(2n+1)/2^(n+1)
Sn=2-(2n+1)/2^n,n为正整数