A Case of Divergence

Solution

The way the problem is posed, the recursive relation is a red herring - a detractor. It is suggesting that, regardless of the choice of $a_1,\,$ $a_n\gt\displaystyle\frac{1}{2016},\,$ for $n\gt 1.\,$ It follows that the series $a_1+a_2+a_3+\ldots\,$ is divergent, so that sooner or later a partial sum will exceed $2017.$

This solves the problem.

Acknowledgment

The problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Leo Giugiuc fast came up with a solution.

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