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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 $m$.

Dr. Wolfgang Hintze
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    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? – Clement C. Nov 30 '19 at 23:05
  • @ 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. – Dr. Wolfgang Hintze Nov 30 '19 at 23:14
  • @ 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? – Dr. Wolfgang Hintze Dec 01 '19 at 08:43
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    I don't know, and cannot speak for the people here. I personally didn't vote either way. – Clement C. Dec 01 '19 at 09:01
  • Well so as to speak, most of the time when we introduce the floor function, the Harmonic numbers come into play, so a closed form expression is much harder or not possible within the scope of current mathematics. Well It's always good so as to hunt these interesting series anyhow. – user712576 Dec 01 '19 at 14:06
  • @ user712576 I agree, and I am a bit surprised that in a Math forum people reject mathematical questions because the proposed problems might have complicated solutions or even no solutions. IMHO this situation should be a pleasant challenge which, as many people know, can give rise at least to interesting side result. – Dr. Wolfgang Hintze Dec 01 '19 at 19:27
  • An interesting problem is worth working on. There doesn't need to be any other reason or context or similar garbage. – marty cohen Dec 01 '19 at 20:20
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    @ 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 https://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). – Dr. Wolfgang Hintze Dec 02 '19 at 12:55

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