若数列cn=(2^n+1)/(2^n-1)求证c2+c3+…+cn＜n+1/3

若数列cn=(2^n+1)/(2^n-1)求证c2+c3+…+cn＜n+1/3
若数列cn=(2^n+1)/(2^n-1)求证c2+c3+…+cn＜n+1/3

若数列cn=(2^n+1)/(2^n-1)求证c2+c3+…+cn＜n+1/3
证：
cn=(2ⁿ+1)/(2ⁿ-1)=(2ⁿ-1+2)/(2ⁿ-1)=1+ 2/(2ⁿ-1)
c2+c3+...+cn
=(n-1)+2[1/(2²-1)+1/(2³-1)+...+1/(2ⁿ-1)]
{1/[2^(n+1) -1]}/[1/(2ⁿ-1)]
=(2ⁿ-1)/(2×2ⁿ-1)
=(2ⁿ -1/2 -1/2)/(2×2ⁿ-1)
=(1/2) -(1/2)/(2×2ⁿ-1)