2002 OIM Problems/Problem 5

Problem

The sequence of real numbers $a1, a2, \cdots$ is defined as:

$a_1 = 56, a_{n+1} = a_n - \frac{1}{a_n}$

for every integer $n \ge 1$ .

Prove that there exists an integer $k$ , $1 \le k \le 2002$ , such that $a_k < 0$ .

~translated into English by Tomas Diaz. orders@tomasdiaz.com

Solution

Notice that every time we apply the recursion, we are essentially subtracting $\frac{1}{a_n}$ from the current term. Clearly, the sequence is decreasing, so for all $n$ , $a_n\le56$ , so $\frac{1}{a_n}\ge\frac{1}{56}$ . Then, after every application of the recursion, the value of $a_i$ will decrease by at least $\frac{1}{56}$ ; clearly, it will eventually reach negative numbers.

~ eevee9406

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2002 OIM Problems/Problem 5

Problem

Solution

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