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.

Welcome to mathimatikoi.org; a site of university mathematics! Enjoy your stay here!

## 16 th Cuban Mathematical Competition of Universities [Problem 5]

- Tolaso J Kos
- Administrator
**Articles:**0**Posts:**844**Joined:**Sat Nov 07, 2015 6:12 pm**Location:**Larisa-
**Contact:**

### 16 th Cuban Mathematical Competition of Universities [Problem 5]

**Imagination is much more important than knowledge.**

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

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.

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.

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