A Floor-Sqrt Recurrence Limit

Problem

A sequence {xn}\{x_n\} satisfies x1=1x_1=1 and xn+1=xn+⌊xn⌋x_{n+1}=x_n+\lfloor\sqrt{x_n}\rfloor. Find

lim⁡n→+∞xnn2.\lim_{n\to+\infty}\frac{x_n}{n^2}.

Answer

Solution

Difficulty8/10
TopicsLimits, calculus, sequences, Estimation, Induction

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