# The sum $\sum_{n=1}^\infty \min_{k\le n}\{\alpha k\}$ for irrational $\alpha$

Let $\alpha$ be an irrational number. For every $n$ let $z_n$ be the integer closest to the number $\alpha n$. Then we can define $$A(\alpha):= \sum_{n=1}^\infty |\alpha n - z_n|.$$

We can also define somewhat similar quantity $$B(\alpha):= \sum_{n=1}^\infty \left(\alpha n - \lfloor \alpha n \rfloor \right).$$

$A(\alpha)$ and $B(\alpha)$ are sums of non-negative numbers, so the values can be real numbers of $+\infty$.

EDIT: In fact, the above sums are clearly both equal to $+\infty$, since $n \alpha$ will be infinitely often arbitrarily close to and integer (and also to a number of the form $k+\frac12$, $k\in\mathbb Z$). See, for example, Multiples of an irrational number forming a dense subset (and several other posts on MSE).

I have missed this obvious fact when posting the question. However the sums $A'(\alpha)$ and $B'(\alpha)$ described below might still be interesting.

I left here the part about $A(\alpha)$ and $B(\alpha)$, since some users already posted some comments about these sums. If I edited the first part of the post away, those comments would not make sense.

We can also make modifications where we replace the sequence with monotone sequences: \begin{align*} A'(\alpha):=& \sum_{n=1}^\infty \min_{k\le n}|\alpha k - z_k|;\\ B'(\alpha):=& \sum_{n=1}^\infty \min_{k\le n} \left(\alpha k - \lfloor \alpha k \rfloor\right).\\ \end{align*}

In some sense, these quantities tell us how far $\alpha$ is from rational numbers. (The monotone versions seem more closely related to question whether $\alpha$ is rational or irrational, since for rational $\alpha$ all but finitely many terms are zero, so the sum must be finite.)

My question is:

• Were the numbers $A'(\alpha)$, $B'(\alpha)$ studied somewhere?
• Can they be calculated for some specific irrational numbers? For example, can we calculate them for $\alpha=\ln 2$, $\alpha=\sqrt2$, $\alpha=e$, $\alpha=\pi$ or some other well-know irrational numbers?
• Do we get $A'(\alpha)=\infty$ or $B'(\alpha)=\infty$ for some numbers? If yes, can such numbers be characterized?
• Are they related to irrationality measure or some other ways to measure how irrational a given number $\alpha$ is?

To explain what lead me to these sums: I was thinking about question whether there is a counterexample to Minkowski's theorem for unbounded sets. (Such question was asked here. In the meantime I learned here that such example cannot be found.) So I was trying to take a line $y=\alpha x$ which does not contain any lattice points. I tried to replace the segments of this line by wider rectangles - this lead me to try to compute how wide rectangles I can add without hitting some lattice point. This is related to the question how far are the points $(n,n\alpha)$ from lattice points.

• You may wish to have a look at Pisot numbers. – A.P. May 2 '15 at 12:41
• @MichaelBurr I meant to evaluate them. (I.e., if they are $+\infty$ for each $\alpha$, that would answer my question.) However, I do not think that $A'(\alpha)$ and $B'(\alpha)$ are always infinite. – Martin Sleziak May 2 '15 at 12:44
• $A$ and $B$ should be infinite for all nonintegral $\alpha$ (they are periodic for $\mathbb{Q}\setminus\mathbb{Z}$). The other two sums are more interesting (indeed) – Michael Burr May 2 '15 at 12:46
• It is not obvious to me that $A(\alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $\alpha$ then $|\alpha n-z_n|$ should behave like $\frac{1}{n}$, leading to a divergent sum. – Jack D'Aurizio May 2 '15 at 13:02
• I bet you can compute these sums for simple quadratic irrationals such as the golden ratio. – Gerry Myerson May 9 '15 at 9:15

## 1 Answer

$$\newcommand{\bivec}{\begin{bmatrix} #1 \\ #2 \end{bmatrix}}$$ Excavating. Still the series for $$A'(\alpha)$$ and $$B'(\alpha)$$ both diverge always.

Let $$\alpha=[a_0;a_1,a_2,\ldots]$$ be the continued fraction expansion of $$\alpha$$, and $$p_n/q_n$$ be its $$n$$-th convergent: $$\bivec{p_{-1}}{q_{-1}}=\bivec{1}{0},\ \bivec{p_0}{q_0}=\bivec{a_0}{1},\ \bivec{p_n}{q_n}=a_n\bivec{p_{n-1}}{q_{n-1}}+\bivec{p_{n-2}}{q_{n-2}}.$$ Then for $$r\geqslant 0$$ and $$q_r\leqslant n we have $$\displaystyle\min_{k\leqslant n}|\alpha k-z_k|=|\alpha q_r-p_r|$$, which gives $$A'(\alpha)=\sum_{n=0}^{\infty}(q_{n+1}-q_n)\cdot|\alpha q_n-p_n|\geqslant\sum_{n=0}^{\infty}\frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$ by "Theorem 5" here. But $$r_n=q_{n+1}/q_n$$ can't converge to $$1$$ because $$r_n=a_{n+1}+1/r_{n-1}$$.

(Therefore $$\dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=\dfrac{r_n-1}{r_n+1}$$ can't converge to $$0$$.)

Thus $$A'(\alpha)=+\infty$$ for any (irrational) $$\alpha$$, and therefore $$B'(\alpha)=+\infty$$ too.