MathDB

3

Part of 1987 Austrian-Polish Competition

Problems(1)

f (x + 1) = f (x) + 1, x_{n+1} = f (x_n), x_0 = a

Source: Austrian Polish 1987 APMC

4/30/2020
A function f:R→Rf: R \to R satisfies f(x+1)=f(x)+1f (x + 1) = f (x) + 1 for all xx. Given a∈Ra \in R, define the sequence (xn)(x_n) recursively by x0=ax_0 = a and xn+1=f(xn)x_{n+1} = f (x_n) for n≥0n \ge 0. Suppose that, for some positive integer m, the difference xm−x0=kx_m - x_0 = k is an integer. Prove that the limit lim⁡n→∞xnn\lim_{n\to \infty}\frac{x_n}{n} exists and determine its value.
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