328 reputation
111
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location Ho Chi Minh City, Vietnam
age
visits member for 2 years, 6 months
seen Dec 6 '13 at 12:04

Jan
20
awarded  Popular Question
Jul
31
awarded  Popular Question
May
23
awarded  Yearling
Dec
22
accepted How many all prime numbers p with length of bits of p = 1024 bits?
Dec
22
asked How many all prime numbers p with length of bits of p = 1024 bits?
Dec
8
answered Show that $\alpha_1u+\alpha_2v+\alpha_3w=0\Rightarrow\alpha_1=\alpha_2=\alpha_3=0$
Dec
7
accepted How to calculate this limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$
Dec
7
accepted Why can $N$ not be the subgroup $\{I, F\}$?
Dec
7
comment Why can $N$ not be the subgroup $\{I, F\}$?
thanks, but what's picture on en.wikipedia.org/wiki/Dihedral_group#Definition???
Dec
7
asked Why can $N$ not be the subgroup $\{I, F\}$?
Dec
7
asked Prove that $f$ is not onto
Feb
15
asked How to calculate this limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$
Dec
28
accepted Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
Dec
28
revised Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
added 226 characters in body
Dec
28
comment Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
I found $\mathcal{O}(S_n) = n^2$. Thus, having $(n-k+1)/k = (n+1)/k -1 \leq n$. ==> $S_n = \sum_{k = 1} ^ {n}n = n^2$. But I cant find $\mathcal{\Omega}(S_n)$, so I cant also find $\mathcal{\Theta}(S_n)$
Dec
28
comment Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
@mt_: I'm sorry. I typed not correct. I editted. Thanks!
Dec
28
revised Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
edited body; edited title
Dec
28
asked Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$
Dec
28
accepted Finding the asymptotics of a summation
Dec
27
accepted Find $a_n$ with $n\geq 0$