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The questions at issue are this and this. If they get no answer even after the end of bounties, might I post to MO?
And how about evaluating $\displaystyle \sum_{m=0}^\infty \sum_{n=1}^\infty \frac{(-1)^mn}{2^n+m}$?

$\begingroup$Why are you interested in this particular series, and what kind of answer would you expect -- a numeric approximation, or what else?$\endgroup$
– Stefan KohlOct 20 '14 at 20:20

$\begingroup$@StefanKohl The three of these series come up in a proof I'm developing, though I'm afraid I don't have the means to work out them. And... well, I would need exact results, but I guess it's asking too much?$\endgroup$
– Vincenzo OlivaOct 20 '14 at 20:26

5

$\begingroup$Taking an arbitrary convergent series, typically its value has no "closed form expression". -- Do you know any reasons why for your series such expressions should exist?$\endgroup$
– Stefan KohlOct 20 '14 at 20:48

$\begingroup$@StefanKohl None besides being part of an expression (whose other terms I'm trying to find on my own) that gives a precise value. At worst, I would not refuse numeric approximations and proofs of rationality or irrationality of the values of the series.$\endgroup$
– Vincenzo OlivaOct 20 '14 at 21:07

$\begingroup$@StefanKohl So? Are they suitable for MO?$\endgroup$
– Vincenzo OlivaOct 21 '14 at 3:20

$\begingroup$The series looks kind of like Lambert series. So maybe you could ask if the series can be expressed in terms of named quantities like that. Did you try Wolframalpha.com?$\endgroup$
– Bjørn Kjos-HanssenOct 21 '14 at 3:56

$\begingroup$@BjørnKjos-Hanssen Yes, and it doesn't even compute $\displaystyle \sum_{n=1}^\infty \frac{n}{2^n} - \sum_{n=1}^\infty \frac{n}{2^n+1}$. Or at least, it will take centuries. (The same goes for the series involving the Lerch transcendent, while the other, which is the subseries in the one displayed in the body of the question, does get computed, but only approximately, and as I said, I would need at least a proof of irrationality or rationality)$\endgroup$
– Vincenzo OlivaOct 21 '14 at 4:27

$\begingroup$@BjørnKjos-Hanssen Truthfully, W|A does compute the equivalent $\displaystyle \sum_{n=1}^\infty \frac{n}{2^n}-\frac{n}{2^n+1}$, but, I would need to do this infinitely many times (the software doesn't understand my infinite sum of infinite sums) and I would have no proof of the (ir)rationality of it.$\endgroup$
– Vincenzo OlivaOct 21 '14 at 5:42

4

$\begingroup$Your double series is convergent but not absolutely convergent, so you have to be careful with your manipulations. Asking for the exact value or rationality of a difficult series might be too difficult (as such things are generally unknown), but if you are willing to accept estimates, you might get an answer. "Too hard for MSE" can mean "impossible" instead of "interesting for MO".$\endgroup$
– Joonas IlmavirtaOct 21 '14 at 8:36

$\begingroup$@JonasIlmavirta I see. I am willing To accept estimates, but the proof of (ir)rationality of the values is something I need. Over all, will my questions be downvoted on MO?$\endgroup$
– Vincenzo OlivaOct 21 '14 at 9:54

$\begingroup$Depends on how you pose your questions. It's hard to tell unless you can give an example of a complete question here (assuming it's not very long). I don't know how I would vote myself without an explicit example. Remark: The @ notification system in comments works only if you spell the name correctly. (I'm not upset by misspelling my name. It just doesn't work, that's all. And it is possible to misspell it into something rude in Finnish...)$\endgroup$
– Joonas IlmavirtaOct 21 '14 at 10:35

$\begingroup$@JoonasIlmavirta Apologies for the misspelling, it is due to using a brand new mobile whose automatic correction truly is a pain in the neck (And I'm Italian). As to the questions... are those on MSE not a good example? I was planning to post exactly them to MO, in a week.$\endgroup$
– Vincenzo OlivaOct 21 '14 at 11:29

1

$\begingroup$Ah, yes, your MSE questions are a good example. Personally, I would probably not vote in either direction. (Since I'm not strongly against such questions.) The questions are not bad, but they seem disconnected from any research problem and I see no reason to believe that anyone could answer them. But I'm not the greatest expert on series, so don't rely on my word too much. Someone else might consider them off-topic. Remember: If you ask, make it clear what you want to know about the series and what would be a useful answer.$\endgroup$
– Joonas IlmavirtaOct 21 '14 at 13:12

8

$\begingroup$There would be less chance of having the question closed if you explain better what the real purpose or motivation of the question is. And it may be that the thing you are really after could be approached in a way different from the need to compute the exact value of certain series which might well be impossible to evaluate.$\endgroup$
– Todd Trimble♦Oct 22 '14 at 13:23

2

$\begingroup$Why are you guys all just solving the question in comments on meta.MO. Shouldn't you just go answer it somewhere?$\endgroup$
– djechlinOct 24 '14 at 9:15

ispossible to misspell it into something rude in Finnish...) $\endgroup$ – Joonas Ilmavirta Oct 21 '14 at 10:35