By trying to compute a few terms first, we find $a_1=1$, $a_2=\frac{1}{2}$, $a_3=\frac{1}{3}$, $a_4=\frac{1} {4}$,$\cdots$. Hence, a natural guess is that $a_n=\frac{1}{n}$. Indeed, this conclusion can be proved by mathematical induction. When $n=1$, $a_1=\frac{1}{1}$. Assume $a_k=\frac{1}{k}$, then when $n=k+1$, $$a_{k+1}=\frac{a_k}{1+a_k}=\frac{\frac{1}{k}}{1+\frac{1}{l}}=\frac{1}{k+1}$$ Thus, by the principle of mathematical induction, we assert $a_n=\frac{1}{n}$.