f(k) = 1/ [√k(k+1) *(√k+√k+1) = (√k+1 - √k) / √k*√(k+1) = 1/√k - 1/√k+1
所以
f(1)+f(2)+.+f(n)
=1-1/√2+1/√2-1/√3+.+1/√n - 1/√(n+1)
=1-1/√(n+1)