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Problem 1297

AIME late
Algebra Difficulty 5.4 Prove it Romanian Mathematical Olympiad · Romania

For each positive integer nn the function fn:[0,n]→Rf_n : [0, n] \to \mathbb{R} is defined by fn(x)=arctg⁡(⌊x⌋)f_n(x) = \arctg(\lfloor x \rfloor). Prove that fnf_n is a Riemann integrable function

and find
lim⁡n→∞1n∫0nfn(x)dx. \lim_{n \to \infty} \frac{1}{n} \int_{0}^{n} f_{n}(x) dx.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

The function fnf_n is locally constant, hence Riemann integrable.

Next, we have
∫0nfn(x)dx=∑i=0n−1∫ii+1fn(i)dx=∑i=0n−1arctan⁡i. \int_{0}^{n} f_{n}(x) dx = \sum_{i=0}^{n-1} \int_{i}^{i+1} f_{n}(i) dx = \sum_{i=0}^{n-1} \arctan i.
Applying Stolz-Cesàro theorem, we obtain
lim⁡n→∞arctan⁡1+arctan⁡2+⋯+arctan⁡nn=lim⁡n→∞arctan⁡n=π2. \lim_{n \to \infty} \frac{\arctan 1 + \arctan 2 + \dots + \arctan n}{n} = \lim_{n \to \infty} \arctan n = \frac{\pi}{2}.

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