A Case of Divergence
Problem
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.
What Is Red Herring
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- Simultaneous Diameters in Concurrent Circles
- An Inequality from the 2015 Romanian TST
- Schur's Inequality
- Further Properties of Peculiar Circles
- Inequality with Csc And Sin
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- A Case of Divergence
- An Inequality for the Cevians through Spieker Point via Brocard Angle
- An Inequality In Triangle and Without
- Problem 3 from the EGMO2017
- Mickey Might Be a Red Herring in the Mickey Mouse Theorem
- A Cyclic Inequality from the 6th IMO, 1964
- Three Complex Numbers Satisfy Fermat's Identity For Prime Powers
- Probability of Random Lines Crossing
- Planting Trees in a Row
- Two Colors - Three Points
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