Recursive sequence

We claim that $a_n = \dfrac{1}{n+1}$ for all $n$ non-negative integers and then prove it by induction.

1) For $n=0$ , $a_n = \dfrac{1}{0+1} = 1$ as given. Therefore, the claim is true for $n=0$ .

2) Assuming that the claim is true for $n$ , then

$\begin{aligned} \quad a_{n+1} & = \frac{a_n}{1+a_n} \\ & = \frac{\frac{1}{n+1}}{1+\frac{1}{n+1}} \\ & = \frac{1}{n+2} \end{aligned}$ .

$\quad$ It is also true for $n+1$ .

Therefore, the claim is true for all $n \ge 0$ .

Now we have, $a_{2016} = \dfrac{1}{2016+1} = \dfrac{1}{2017}$

$\Rightarrow A+B = 1 + 2017 = \boxed{2018}$

If we define $b_n = a_n$ , $b_0 = 1$ and $b_n = 1 + b_{n-1}$ . Then it is easy to prove that $b_n = n+1$ , so $a_n = \frac{1}{n+1}$ .

Finally, $a_n = \frac{1}{2017}$ and $A+B=2018$ .

$n \geq1$ . In question it is given that $n>1$ . Also by easy analysis of first few terms one can easily make out that $a_n$ = ${1}/{1+n}$ . Hence $a_{2016}$ = ${1}/{2017}$ . Therefore answer = $\boxed{2018}$