Reciprocal fraction

For $k > 1$ , we have

$\begin{aligned} \frac{1}{a_{k+1}} &= \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k} \\ \frac{1}{a_{k+1}} &= \frac{1}{a_k} + \frac{1}{a_k} \\ \frac{1}{a_{k+1}} &= \frac{2}{a_k} \\ \frac{a_{k+1}}{a_k} &= \frac{1}{2} \end{aligned}$

This is simply the Geometric Progression! Each term is divided by 2, hence we have $a_{2016} = \frac{3}{2^{2014}}$ .

We have $a_1 = 3$ .

First, we have to find the value of $a_2 , a_3 , a_4 , ......$ so we can determine the sequence.

For $k=1$ , we have

$\frac{1}{a_2} = \frac{1}{a_1}$

So we have $a_1 = a_2 =3$ .

For $k=2$ , we have

$\frac{1}{a_3} = \frac{1}{a_1} + \frac{1}{a_2}$

Then we have $\frac{1}{a_3} = \frac{2}{3}$

For $k=3$ , we have

$\frac{1}{a_4} = ( \frac{1}{a_1} + \frac{1}{a_2} ) + \frac{1}{a_3}$

$\frac{1}{a_4} = 2 \times \frac{1}{a_3}$

$\frac{1}{a_4} = \frac{4}{3}$

So we can conclude that $\frac{1}{a_n} = \frac{2^{n-2}}{3}$

Now, $\frac{1}{a_{2016}} = \frac{2^{2014}}{3}$

$a_{2016} = \frac{3}{2^{2014}}$