Reputation
4,474
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
2 9 47
Impact
~49k people reached

Feb
5
awarded  Notable Question
Feb
2
comment Liftings of curves $u\cdot v$ and $v\cdot u$ with respect to the sine covering map.
@ts375_zk26: Yes, $u\cdot v$ means the curve obtained by first tracing $u$ and then $v$.
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@RRL: I know $t_N$ is well-defined because the series it represents converges. Writing $\sum_{n=1}^{\infty} |a_n|$ inside $t_{N}$ is just algebraic manipulation. I am not assuming that $\sum_{n=1}^{\infty} |a_n|$ is finite there. I think there ought to be a better write-up of this approach that doesn't make it seem circular.
Jan
26
revised Show that $\sum_{n=0}^\infty |a_n|<\infty.$
corrected a mistake
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@RRL: I say $t_N$ is well-defined because we have proved it to be! I don't think there is a circular logic here, but perhaps my presentation is poor. Also, thanks for the catch. It should be $\sum (a_n - |a_n|) \leq \sum 2a_n$ because if $a_n$ is positive, then $a_n-|a_n|=0$, and if $a_n$ is negative, then $a_n-|a_n|=2a_n$. Thus, $\sum (a_n-|a_n|)\leq \sum 2a_n$. I will edit the answer accordingly.
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@EpsilonDelta I have now edited my answer. Thanks for the heads-up, and I hope the argument is simpler now.
Jan
26
revised Show that $\sum_{n=0}^\infty |a_n|<\infty.$
the argument is now simpler
Jan
26
answered Show that $\sum_{n=0}^\infty |a_n|<\infty.$
Jan
25
comment Every group is the disjoint union of generator of cyclic subgroup of G
@Alan Wang: Yes, exactly!
Jan
25
answered Every group is the disjoint union of generator of cyclic subgroup of G
Jan
22
awarded  Revival
Jan
22
comment Elementary proof of the prime number theorem?
I don't understand what you mean by "since this grows in a certainly nonlinear way, we can write …" Could you be more rigorous?
Jan
17
answered Stuck proving that if $m$ and $n$ are perfect squares. Then $m+n+2\sqrt{mn}$ is also a perfect square.
Jan
15
comment An identity involving $[\sigma(n)]^2$
Excellent! Thanks a lot. It is good that I also learned this nice little fact $m\mid n^2$ $\Rightarrow$ $\gcd(\frac{n^2}{m}, m) \mid n$ from this post.
Jan
15
comment An identity involving $[\sigma(n)]^2$
Thanks. Note that proof 2 is included in the original post :)
Jan
15
accepted An identity involving $[\sigma(n)]^2$
Jan
12
comment An identity involving $[\sigma(n)]^2$
Even though it only gives the proof for prime power case, this is a cool approach. Thanks!
Jan
12
comment An identity involving $[\sigma(n)]^2$
@Thomas Thanks! I knew this operation is called Möbius convolution but didn't realize that it preserves multiplicative property. So that verifies my proof. I'd still love to see an alternative approach (say with double counting).
Jan
12
asked An identity involving $[\sigma(n)]^2$
Dec
26
answered How does the Schwarz inequality prove the following?