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Here is a seemingly minor modification of the problem Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$. We change the exponent of $k$ from $\frac{1}{2}$ to $\frac{2}{3}$ and ask the same

Question: is there a closed form of this sum

$$s =\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\left\lfloor k^{\frac{2}{3}}\right\rfloor \simeq 0.8129$$

This slight change makes the problem more complicated because one obvious approach is trying to get rid of the floor function by setting $k=n^3+j$ with $0\le j \le 3n(n+1)$. But in contrast to the previous cases we cannot assert anymore that $\left\lfloor (n^3+j)^{\frac{2}{3}}\right\rfloor = n^2$. This prevents simplifying the floor function which had worked for powers of $k$ of the type $\frac{1}{m}$ with a positive integer $m$.

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    $\begingroup$ No offense, but can I ask you why you are systematically going through all combinations of floors and powers possible? Is there a deeper reason? $\endgroup$
    – Clement C.
    Commented Nov 30, 2019 at 23:05
  • $\begingroup$ @ Clement De nada. This is a completely valid question. I was certainly not going through all combinations, I just try to explore the topic a bit. It is just for fun, there is no "deeper" reason. I also liked problems with integrals over functions of the fractional part of the integration variable. Both types are interesting (at least to me) because they leave the realm of continuous functions. $\endgroup$
    – Dr. Wolfgang Hintze
    Commented Nov 30, 2019 at 23:14
  • $\begingroup$ @ Clement I am noticing a series of 5 downvotes. From 1 to 0. I wonder if those people have ever looked at the hundreds of problems concerning harmonic numbers (where I have also participated) without pointing to "deeper reasons". Marty Cohen has has recently even been put on hold ... kind of floor-phobia here? $\endgroup$
    – Dr. Wolfgang Hintze
    Commented Dec 1, 2019 at 8:43
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    $\begingroup$ I don't know, and cannot speak for the people here. I personally didn't vote either way. $\endgroup$
    – Clement C.
    Commented Dec 1, 2019 at 9:01
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    $\begingroup$ @ marty cohen Exactly! Thank you for your statement (which I errenuously thought to be consensus in a Math forum :-(). You might be interested in my questions to the "dear downvoters" in math.stackexchange.com/q/3456477/198592. Because the problem appears frequently and I am annoyed independently of who's the victim, I'm thinking about putting forward a motion in Meta which for downvoter requests a mandatory user name and a reasoning. This could prevent what could be called "sniper" mentality (inflicting damage invisibly from a secure post). $\endgroup$
    – Dr. Wolfgang Hintze
    Commented Dec 2, 2019 at 12:55

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