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For each $k=1, 2, \dotsc$ , n let $S_k = \left\lfloor \frac{x+0}{k}\right\rfloor + \left\lfloor \frac{x+1}{k} \right\rfloor + \left\lfloor\frac{x+2 }{k} \right\rfloor + \dotsb + \left\lfloor \frac{x+(k-1) }{k}\right\rfloor.$

Show that $S_1+S_2+\cdots +S_n=n\lfloor x \rfloor$.

Here is what can be done:

$S_1=\lfloor x\rfloor$

$S_2=\left\lfloor\frac{x+1}{2}\right\rfloor=\left\lfloor\frac{x}{2}+\frac12\right\rfloor \rightarrow y=\frac{x}{2},\rightarrow \left\lfloor y+\frac12\right\rfloor=\left\lfloor2\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\lfloor=\lfloor x\rfloor-\left\lfloor\frac{x}{2}\right\rfloor$

Here $S_2$ is solved by using the equality $\left\lfloor y+\frac12\right\rfloor=\lfloor 2y\rfloor -\lfloor y\rfloor$.

If we get that $S_3=\lfloor x\rfloor +\lfloor \frac{1}{2}\rfloor$ then summing it with $S_2$ would give $\lfloor2 x\rfloor$, we should be on the way to the answer $S_1+S_2+S_3=3\lfloor x\rfloor$. But which equality of floor functions can I use to find show that $\left\lfloor\frac{x+2}{3}\right\rfloor=\lfloor x\rfloor$?

Thanks

Luthier415Hz
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    This might help: https://en.wikipedia.org/wiki/Hermite%27s_identity – Martin R Aug 02 '23 at 08:58
  • Thanks Martin, I wasn't aware of this link. – Luthier415Hz Aug 02 '23 at 08:59
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    See also https://math.stackexchange.com/q/522709/42969 (which, I think, qualifies as a duplicate target for your question) – found with Approach0 – Martin R Aug 02 '23 at 09:00
  • Can't see how $\frac{x+k-1}{k}$ can be rewritten to anything familiar but to $\frac{x}{k}+\frac{k-1}{k}$, and how that is related to Hermite's sum – Luthier415Hz Aug 02 '23 at 09:04
  • @MartinR there is a missing proof or calculation in that post you shared. – Luthier415Hz Aug 02 '23 at 09:07
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    Your $S_2$ is wrong. The actual value of $S_2$ is $\lfloor x/2 \rfloor + \lfloor (x + 1)/2\rfloor$ and not $\lfloor (x + 1)/2\rfloor$. Hence, $S_1 + S_2 = \lfloor x \rfloor + \lfloor x/2 \rfloor + \lfloor x \rfloor - \lfloor x/2\rfloor = 2\lfloor x \rfloor$. This should give you a good idea on how to do this question by induction. – Alias K Aug 02 '23 at 09:32
  • @AliasK thanks for the tip! Regarding $S_2$ did you use that same equality I gave (with the $y$ variable)? – Luthier415Hz Aug 02 '23 at 10:31
  • @AliasK shouldn't it rather be $S_2=\lfloor x/2+\frac{1}{2}\rfloor\rightarrow \lfloor x\rfloor - \lfloor\frac{x}{2}\rfloor$? – Luthier415Hz Aug 04 '23 at 09:09

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