P(A\leq k)=\sum_{i=1}^k P(A=i)=\sum_{i=1}^k {\frac {1}{i(i+1)}=\sum_{i=1}^k ({\frac {1}{i}}-{\frac {1}{i+1}})=1-{\frac {1}{k+1}}={\frac {k}{k+1}}