Answer
$a_n=\frac{F_n}{F_{n+1}}$, see explanations.
Work Step by Step
Step 1. Given $a_{n+1}=\frac{1}{1+a_n}$ and $a_1=1$, we have $a_2=\frac{1}{2}$, $a_3=\frac{2}{3}$, $a_4=\frac{3}{5}$, $a_5=\frac{5}{8}...$
Step 2. Recall the Fibonacci numbers: $1,1,2,3,5,8,...$ and compare with the above sequence, we have $a_1=\frac{F_1}{F_2}$, $a_2=\frac{F_2}{F_3}$, $a_3=\frac{F_3}{F_4}$, $a_4=\frac{F_4}{F_5}$, $a_5=\frac{F_5}{F_6}...$, $a_n=\frac{F_n}{F_{n+1}}$
Step 3. Recall the property of Fibonacci sequence, we have: $F_{n}=F_{n-1}+F_{n-2}$ and $F_1=F_2=1$
Step 4. Prove the statement is true when $n=1$: $a_1=\frac{F_1}{F_2}=1$, thus it is true for $n=1$.
Step 5. Assume the statement is true when $n=k$: we have $a_k=\frac{F_k}{F_{k+1}}$
Step 6. Prove it is true for $n=k+1$: $LHS=a_{k+1}=\frac{1}{1+a_k}=\frac{1}{1+\frac{F_k}{F_{k+1}}}=\frac{F_{k+1}}{F_{k+1}+F_k}=\frac{F_{k+1}}{F_{k+2}}$
With $n=k+1$, we have $RHS=\frac{F_{k+1}}{F_{k+2}}$. Thus, the statement is also true for $n=k+1$
Step 7. Conclusion: with mathematical induction, we have proved that the statement is true for all natural numbers $n$.
