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 Posted: Thu Apr 27, 2017 8:31 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 836
Location: Larisa
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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 Posted: Sat May 26, 2018 1:00 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 146
Location: Melbourne, Australia
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.

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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