71 reputation
6
bio website douban.com/people/project309
location Chengdu, Sichuan, China
age 27
visits member for 3 years, 5 months
seen Nov 24 at 2:33

I am a Chinese undergraduate student who love mathematics.


Jan
4
awarded  Nice Question
Jan
10
awarded  Announcer
Jun
28
comment Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
OK. I understand now.
Jun
28
comment Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
$(n+1)^3 = \sum_{k=0}^n (k+1)^3 - k^3$ ? $(n+1)^3 = \sum_{k=0}^n (k+1)^3 - \sum_{k=0}^n k^3$
Jun
27
awarded  Supporter
Jun
27
comment Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
@Beni Bogosel: If you understand Chinese, you can see link. In fact, I have asked on link. Sorry, I am a Chinese. My English is not very good.
Jun
27
awarded  Scholar
Jun
27
comment Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
@TonyK: This problem and Diophantine approximation theory are related. I never studied the Diophantine approximation theory, so to ask next.
Jun
27
accepted Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
Jun
27
comment Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
@mac:I am a Chinese, my English is not very good. If you understand Chinese, you can see link. Thank you.
Jun
27
comment Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
@mac:Let $F=\sqrt[x]{|\sin x|}$, if $x=2k\pi$, $F=0$. If $x=(2k+1/2)\pi$, $F=1$. So $\lim_{x \to \infty} \sqrt[x]{|\sin x|}$ does not exist.
Jun
27
awarded  Student
Jun
27
revised Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
edited tags
Jun
27
asked Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$
Jun
27
awarded  Autobiographer