f(1/n) = 1/2 + log₂ [(1/n)/(1-1/n)]
= 1/2 + log₂ 1/(n-1)
= 1/2 +log₂ 1 - log2 (n-1)
f(2/n) = 1/2 +log₂ [(2/n)/(1 -2/n)]
= 1/2 + log₂ [2/(n-2)]
= 1/2 + log₂ 2 - log₂(n-2)
f(3/n) = 1/2 + log₂3 - log₂(n-3)
……
f((n-1)/n) = 1/2 + log₂(n-1) + log₂1
Sn= n/2 【后面的log全部相互抵消了】