4
$\begingroup$

For any real number $a$ and a positive integer $n$, there is a concise formula to calculate

$$a + 2a + 3a + \cdots + na = \frac{n(n+1)}{2} a.$$

The proof for the same is given in Mathematical literature.

Is there any such formula to calculate:

$$\lfloor a\rfloor + \lfloor 2a\rfloor + \lfloor 3a\rfloor + \cdots + \lfloor na\rfloor $$

and

$$\lceil a\rceil + \lceil 2a\rceil + \lceil 3a\rceil + \cdots + \lceil na\rceil $$

for any whole number $n$ and $ 0 < a < 1$ ? Also, provide the proof for the same.

  • 1
    Have you tried calculating any, looking for patterns?2012-10-05
  • 0
    Yes, tried calculating for both things. There is a pattern for particular numbers like 0.2, 0.4, 0.8. But, how to generalize the pattern for any real number like 0.39856, 0.0009843, etc? Of course, finding general patterns helps in providing a formula. There doesn't seem having a general pattern among all real numbers.2012-10-05
  • 0
    For rational $a$, your sums come up in some of the early proofs of quadratic reciprocity --- see, e.g., Eisenstein's proof http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity2012-10-05

2 Answers 2

6

There seems no exact formula for the sum, considering that the sum depends highly sensitively on the fraction part of $a$. But we can give an asymptotic formula:

Case 1. If $a$ is rational, write $a = p/q$ where $p$ and $q$ are positive coprime integers. Then $kp \ \mathrm{mod} \ q$ attains every value in $\{0, 1, \cdots, q-1\}$ exactly once whenever $k$ runs through $q$ successive integers. Thus if we write $n = mq + r$,

$$ \begin{align*} \sum_{k=1}^{n} (ka - \lfloor ka \rfloor) &= \sum_{k=1}^{mq} (ka - \lfloor ka \rfloor) + \sum_{k=1}^{r} (ka - \lfloor ka \rfloor) \\ &= \frac{m(q-1)}{2} + O(1) = \frac{n(q-1)}{2q} + O(1). \end{align*}$$

This gives

$$ \sum_{k=1}^{n} \lfloor ka \rfloor = \frac{1}{2}n\left(n+\frac{1}{q}\right) + O(1). $$

Case 2. If $a$ is irrational, then the fractional parts $\langle ka \rangle := ka - \lfloor ka \rfloor$ is equidistributed on $[0, 1]$ by Weyl's criterion. Thus $$ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \langle ka \rangle = \int_{0}^{1} x \, dx = \frac{1}{2} \quad \Longrightarrow \quad \sum_{k=1}^{n} \langle ka \rangle = \frac{n}{2} + o(n) $$ and we have $$ \sum_{k=1}^{n} \lfloor ka \rfloor = \frac{n^2}{2} + o(n). $$

  • 0
    by user [HKedia](https://math.stackexchange.com/users/453935/hkedia): I am sorry, I dont have enough reputations to comment. @Sangchul Lee what is meant by o(n) in your irrational derivation?2017-06-10
  • 0
    @miracle173 It means a quantity which grows slower than $n$. Formally, if $a_n=o(n)$, then $a_n/n\to0$ as $n\to\infty$.2017-06-10
6

There is a closed form solution to the sum $\sum_{0\le k