2015
Problem - 2609
Define the sequence $a_i$ as follows: $a_1 = 1, a_2 = 2015$, and $a_n =\frac{na_{n-1}^2}{a_{n-1}+na_{n-2}}$ for $n > 2$. What is the least $k$ such that $a_k < a_{k-1}$?
The recursion is equivalent to $$\frac{a_{n-1}}{a_n} = \frac{a_{n-2}}{a_{n-1}} + \frac{1}{n} = \frac{1}{2015}+\sum_{i=3}^n\frac{1}{i}$$
The first $k$ for which $\frac{a_{k-1} }{a_k} > 1$ occurs when $k = 7$ by a simple computation of the sum.