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 Mar2 awarded Popular Question Nov25 awarded Notable Question Sep10 awarded Popular Question Jul2 awarded Curious Jan20 awarded Popular Question Jul31 awarded Popular Question May23 awarded Yearling Dec22 accepted How many all prime numbers p with length of bits of p = 1024 bits? Dec22 asked How many all prime numbers p with length of bits of p = 1024 bits? Dec8 answered Show that $\alpha_1u+\alpha_2v+\alpha_3w=0\Rightarrow\alpha_1=\alpha_2=\alpha_3=0$ Dec7 accepted How to calculate this limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$ Dec7 accepted Why can $N$ not be the subgroup $\{I, F\}$? Dec7 comment Why can $N$ not be the subgroup $\{I, F\}$? thanks, but what's picture on en.wikipedia.org/wiki/Dihedral_group#Definition??? Dec7 asked Why can $N$ not be the subgroup $\{I, F\}$? Dec7 asked Prove that $f$ is not onto Feb15 asked How to calculate this limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$ Dec28 accepted Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$ Dec28 revised Finding the asymptotics of a summation $\sum_{k=1}^{n}\frac{n-k+1}{k}$ added 226 characters in body Dec28 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 $\mathcal{\Theta}(S_n)$ Dec28 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!