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 Apr18 awarded Autobiographer Dec26 awarded Popular Question Oct22 awarded Critic Oct21 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. Aug7 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! Aug7 accepted Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Aug7 comment Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Alright, this makes a lot of sense. Very nice solution! Aug6 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. Aug5 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? Aug5 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. Aug5 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... Aug5 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. Aug5 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? Aug5 asked Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ Jul2 awarded Curious Aug25 comment Finding a paper by John von Neumann written in 1951 Done, thanks again! Aug25 accepted Finding a paper by John von Neumann written in 1951 Aug25 comment Finding a paper by John von Neumann written in 1951 Awesome, thanks! Aug25 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". Aug25 comment Finding a paper by John von Neumann written in 1951 Brilliant, thanks Michael. If you'll post your comment as an answer I'll accept it; otherwise I'll mark this one as the accepted answer. How did you find it? Did you directly think of the BNF?