i am a student from china and i am not good at english.

you can let b(i)=1/1+a(i) then everything is ok

Note that $$ \frac{a_m}{\prod_{k=1}^m(1+a_k)}=\frac{1+a_m}{\prod_{k=1}^m(1+a_k)}-\frac{1}{\prod_{k=1}^m(1+a_k)}=P_{m-1}-P_m, $$where $P_m=\frac{1}{\prod_{k=1}^m(1+a_k)}$ (with $P_0=1$). Hence, we need to prove that $$ \sum_{m=1}^n(P_{m-1}-P_m)\leq\frac{1}{2^n}, $$or $$ 1-P_n\leq\frac{1}{2^n}, $$which is equivalent to $$ \prod_{k=1}^n(1+a_k)\leq 2^n. $$Finally, the last inequalty is an application of AM-GM for numbers $\{1+a_k\}_{k=1}^n$.