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I've seen a couple posts recently about finding closed forms of or calulating $ \sum^n_k \lfloor c \cdot k \rfloor $ for any real $c>0$. And I was wondering about bounds we can give to this expression. Clearly since, $\lfloor ck \rfloor \le ck$, we know that $\sum_k^n \lfloor ck \rfloor \le \sum_k^n ck $. Also $ck-1 < \lfloor ck \rfloor$ so we can write,

$$ \sum_{k=a}^n [ck-1] < \sum_{k=a}^n \lfloor ck \rfloor \le \sum_{k=a}^n ck $$ $$ -\frac{1}{2} (a-n-1)(ac + cn - 2) < \sum_{k=a}^n \lfloor ck \rfloor \le -\frac{1}{2} c (a - n - 1)(a+n) $$

Can we close these bounds in further? Would the average of the bounds, $ \frac{(-1/2)(a-n-1)((ac+cn-2) + c(a+n))}{2} = \frac{(a-n-1)(ac+cn-1)}{-2}$, serve as a good approximation?

Example: If we use the above estimation and let $a=1$, $n=98$, and $c=\sqrt{3}$ we get $\text{actual} = 8353$ and $\text{estimate}=8353.1784675166$, which is only a $0.0021365\%$ error. Another example: let $a=1$, $n=100$, and $c=\pi$ we get $\text{actual}=15815$ and $\text{estimate}=15815.042900628456$, which is $0.0002712654344341099\%$ error.

Such low error percentages to me indicate that this is a viable approximation. In addition, after running a couple hundred test cases it seems that $\lfloor \text{estimate} \rfloor = \text{actual}$. Is this the case? If so, how would one show it?

Dando18
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  • The sequence $A_n = \frac{c}{2}n(n+1) - \frac{n}{2}$ is a good approximation to $V_n = \sum_{k=1}^n\lfloor ck \rfloor$. However, it is not easy to bound the error $\epsilon_n = |V_n-A_n|$. $V_n$ can be different from both $\lfloor A_n \rfloor$ and $\lceil A_n \rceil$. e.g. for $c = \sqrt{3}$, $V_{44521} = 1716582083$ but $A_n \approx 1716582085.058...$, $V_{8636941} = 64602683560348$ but $A_{8636941} \approx 64602683560351.058$. $\epsilon_n$ is usually small but occasionally we see an $\epsilon_n > 1$. The largest error $\max_{k\le n}\epsilon_k$ also seems to grow without bound. – achille hui Jun 03 '17 at 22:53
  • @achillehui Is there a better approximation? Can we show that $\epsilon_k$ grows without bound? – Dando18 Jun 04 '17 at 17:45
  • There may be better approximation but I don't know anything about that. – achille hui Jun 04 '17 at 17:59
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    @achillehui The error is $O(n^{1-\frac1{\mu-1}+\epsilon})$ where $\mu$ is the irrationality measure of $c$. So, in case with $c=\sqrt 3$, $\mu=2$. – Sungjin Kim Jun 08 '17 at 03:14
  • @i707107 interesting, thanks for the info. – achille hui Jun 08 '17 at 04:52
  • @i707107 thanks, could you link to a source for that? I'm curious. – Dando18 Jun 08 '17 at 10:49
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    It can be done by the method of @stewbasic as in this answer: https://math.stackexchange.com/questions/2239269/does-the-series-sum-n-1-infty-frac-sin-cosnn-converge – Sungjin Kim Jun 08 '17 at 11:41
  • @Dando18 A good text on this is "Kuipers & Niederreiter, Uniform Distribution of Sequences" – Sungjin Kim Jun 08 '17 at 21:05

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