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 Jul 11 awarded Popular Question Apr 18 awarded Autobiographer Dec 26 awarded Popular Question Oct 22 awarded Critic Oct 21 comment Dividing by 2 numbers at once, what is the answer? -1. Quoting wikipedia: "For example, subtraction and division, as used in conventional math notation, are inherently left-associative.". The notation 4/1/5 is clear and well-defined. Aug 7 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Very interesting calculation. I'm accepting the simpler answer, but this is indeed an interesting case where the small trends add up. Nice one! Aug 7 accepted Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Aug 7 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Alright, this makes a lot of sense. Very nice solution! Aug 6 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ The result look right, but including terms beyond the first one in the equivalent is useless: since $2^{n+1}/n$ dominates, you could add any constant instead of $2$, and the result would still be valid. Aug 5 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Thanks! I find the "This holds for every ... hence" line a bit confusing though :/ Once you multiply both sides by $n/2^n$ and look at $u \to 1$, doesn't the upper bound reduce to $2n+2$ instead of just $2+\varepsilon_n$? Or did I miss something? Aug 5 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Well, you have a definite integral here. It looks a bit like writing $\forall x, \forall x, x-x=0$, which might be formally valid, but is still confusing. Aug 5 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ I didn't downvote, but plotting suggests that this is wrong, and since you did not explain how you derived the result is hard to gain anything from this answer... Aug 5 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Side note: You seem to be using $n$ as both a bound variable under the integral, and unbound outside; that's somewhat confusing. Aug 5 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Interesting. Do you have a proof, or hints of how you derived this? Aug 5 asked Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Jul 2 awarded Curious Aug 25 comment Finding a paper by John von Neumann written in 1951 Done, thanks again! Aug 25 accepted Finding a paper by John von Neumann written in 1951 Aug 25 comment Finding a paper by John von Neumann written in 1951 Awesome, thanks! Aug 25 comment Which is the “fastest” paper-pencil method to compare $\sqrt[17]{6}$ and $\sqrt[16]{4}$? Note that you were rather lucky here, since sour simplification (using $3^{12} > 2^{12}$) wouldn't always yield a result. Consider for example $3^{13}$ and $2^{14}$ : $3^1 < 2^2$, and yet $3^{13} > 2^{14}$. Thus writing "Now $3^{12}>2^{12}$ and so you need to compare (...)" is not really rigorous (it implies it is necessary, while it is just sufficient); instead I think you should write something like "Now $3^{12}>2^{12}$ and so proving $3^{4}=81 > 2^{6}=64$ would be sufficient".