Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it 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
limn1n0nfn(x)dx. \lim_{n \to \infty} \frac{1}{n} \int_{0}^{n} f_{n}(x) dx.

Solution

The function fnf_n is locally constant, hence Riemann integrable.

Next, we have
0nfn(x)dx=i=0n1ii+1fn(i)dx=i=0n1arctani. \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
limnarctan1+arctan2++arctannn=limnarctann=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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