# 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

- On the Difference of Areas
- Area of the Union of Two Squares
- Circle through the Incenter
- Circle through the Incenter And Antiparallels
- Circle through the Circumcenter
- Inequality with Logarithms
- Breaking Chocolate Bars
- Circles through the Orthocenter
- 100 Grasshoppers on a Triangular Board
- 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
- Area Inequality in Trapezoid
- Triangles on HO
- From Angle Bisector to 120 degrees Angle
- 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

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