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16 th Cuban Mathematical Competition of Universities [Problem 5]

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Tolaso J Kos
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16 th Cuban Mathematical Competition of Universities [Problem 5]

#1

Post by Tolaso J Kos » Thu Apr 27, 2017 8:31 pm

Let $\alpha \in \mathbb{R} \setminus \mathbb{Z}$ and let us denote with $\lfloor \cdot \rfloor$ the floor function. Prove that the series

$$\mathcal{S}= \sum_{n=1}^{\infty} \left(\alpha-\frac{\lfloor n\alpha \rfloor}{n}\right)$$

diverges.
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Re: 16 th Cuban Mathematical Competition of Universities [Problem 5]

#2

Post by Riemann » Sat May 26, 2018 1:00 pm

Greetings,

We are focusing on the $\alpha$' s lying in the interval $(0, 1)$. That is because each term of the series is $1$ periodic. Let $\mathbb{Z} \ni k >0$ and let $n$ be the maximal integer for which it holds

\[k-1 <n\alpha < k\]

Since it holds that $\left \{ n \alpha \right \} \geq 1-\alpha$ as well as $n \leq \frac{k}{\alpha}$ we deduce that the series has at least one term of the form $\displaystyle \frac{\alpha\left ( 1-\alpha \right )}{k}$. Since for every positive integer $k$ we have one such term , we conclude that the series diverges.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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