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  • 45 votes cast
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".
Aug
25
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?