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It is well-known that $$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots = \ln 2 $$ Hence $$\frac{x}{1}-\frac{x}{2}+\frac{x}{3}-\frac{x}{4}+\cdots= x\ln 2 $$ However, consider $f(x)$, where $$f(x)=\left\lfloor\frac{x}{1}\right\rfloor-\left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor-\left\lfloor\frac{x}{4}\right\rfloor+\cdots $$ and $\lfloor t\rfloor$ denotes the floor function. Can anyone determine and prove the asymptotic behaviour of $f(x)$? Maybe even a good error term as well or even the second term in the asymptotic expansion? I'm guessing that $$ f(x) \sim x \ln 2$$ but cannot prove it.

Edit: I believe I may have solved my own problem. The infinite sum is equal to the limit of the partial sum as the amount of terms tends to infinity. The first $n$ terms can be approximated by dropping the floor function signs at a cost of an error of $O(n)$, leaving one with $$\frac{x}{1}-\frac{x}{2}+\frac{x}{3}-\frac{x}{4}+\cdots + (-1)^{n+1}\frac{x}{n} + O(n)$$ The alternating sum multiplied by $x$ is almost $ x \ln 2$ and the error in approximating the partial sum as the infinite sum is $O(\frac{1}{n})$, since the series is alternating. Hence, the error introduced by approximating $$ \frac{x}{1}-\frac{x}{2}+\frac{x}{3}-\frac{x}{4}+\cdots + (-1)^{n+1}\frac{x}{n}$$ as $x \ln 2$ is $O(\frac{x}{n})$ and so the nth partial sum is equal to $$ x \ln2 + O\left(\frac{x}{n}\right) + O(n)$$ But $n$ is independent of $x$ and so we minimise the error by letting $n=\sqrt x$ giving the final result $$f(x)= x \ln2 + O(\sqrt x) $$ I believe I have made no mistake in this solution.

Antonio Vargas
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Asier Calbet
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  • For $x\in \mathbb{R} $ there exists $N\in \mathbb{N}$ such that for $n\geq N$ , $[\dfrac{x}{n}]=0$, then $f(x)=\displaystyle\sum_{k=1}^N(-1)^{k-1}[\dfrac{x}{k}]$ – Hamou Jul 25 '14 at 00:15
  • Correct. Doesn't yield an asymptotic, though. – Asier Calbet Jul 25 '14 at 00:17
  • One can write $\lfloor x \rfloor$ and $\left\lfloor \dfrac x n \right\rfloor$. ${}\qquad{}$ – Michael Hardy Jul 25 '14 at 00:27
  • What is the Latex command for it? – Asier Calbet Jul 25 '14 at 00:28
  • \lfloor and \rfloor – rogerl Jul 25 '14 at 01:07
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    Related: http://math.stackexchange.com/questions/115824/rounding-is-asymptotically-useless – Aryabhata Jul 25 '14 at 03:08
  • I suggest you move your solution as an anwer. – leonbloy Jul 25 '14 at 14:51
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    I think there's a flaw in this argument, but the link Aryabhata provides is helpful. The problem is that $f(x)$ is exactly equal to the $x$th partial sum, but you've only estimated the $\sqrt{x}$th partial sum. This leaves a third error term that is more subtle to estimate since it has $O(x)$ terms in it. But I suspect Dirichlet hyperbola method can be brought to bear on both the odd and even terms to bring it down to $O(\sqrt{x})$ as well. – Erick Wong Jan 02 '15 at 16:20

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I can't leave comments (not enough reputation), so I'm writing here. If I understand you correctly (which I'm not sure I am), I believe your solution in incorrect. For example, for positive integers it seems that $f(2n-1)=f(2n)=n$ (e.g. $f(3)=f(4)=2$), so it can't be true that $f(x)$ is asymptotically $x\ln2$. I think that a useful observation is that generally, for an integer $n$, we get $f\left( n \right)=\sum\limits_{k=1}^{n}{{{\left( -1 \right)}^{k-1}}\left\lfloor \frac{n}{k} \right\rfloor }$ (later terms are all zeros). So the number of nonzero terms in the series is a function of $x$ (and specifically, for every $x$ we get a finite sum - so "taking a limit" becomes trivial). I'm guessing (just guessing) that for positive reals, $f(x)$ is probably (identically) the step function that corresponds to the restriction of $f(x)$ to the integers.

Borbei
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