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Dec
6
comment Negative binomial distribution - sum of two random variables
If $X\sim NB(r,p)$, then $X=k$ means $k$ is the time of $r$-th success. The geometric random variable gives the first time of success.
Dec
6
answered Negative binomial distribution - sum of two random variables
Nov
30
comment Prove that $ AA^T=0\implies A = 0$
Your problems on title and the content are different.
Nov
26
comment Asymptotics for the Alternating Mertens Function
So, for any $K>0$, it is bounded above by $n/(\log n)^K$.
Nov
26
comment Asymptotics for the Alternating Mertens Function
See this: math.stackexchange.com/questions/332832/…
Nov
13
comment Limit of the sequence $(\sin n)^{n}$
$(\sin n)^{n^2}$ diverges as well.
Nov
3
comment If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also exists
Why do you use Jensen's inequality, what about Cauchy-Schwarz?
Oct
15
comment $P(AB=BA)$ , $A,B\in M_{3x3}(\mathbb Z/p\mathbb Z)$
They are THE solutions only if the minimal polynomial of $A$ coincides with the characteristic polynomial.
Oct
13
awarded  Nice Answer
Oct
9
comment An exotic sequence
I now see that your solution implies that the irrational number (it might be even transcendental) $\arctan(\sqrt 7 )/\pi$ has a finite irrationality measure.
Oct
8
comment An exotic sequence
I see. The last line is essentially from $\sin x \sim x$, right?.
Oct
8
comment An exotic sequence
I think the last line should be $|\arg a^k \pm i\pi/2| > (\textrm{constant}\cdot k)^{-\mu}$.
Oct
8
comment Show $\sum_n \left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}} \right)^n$ converges for $\epsilon>0$.
Comparison test.
Oct
7
comment An exotic sequence
That number $2\arctan(\sqrt 7 )/\pi$ is an irrational number.
Oct
7
comment Proving that $\left|\Re\left( \frac{1+i\sqrt{7}}{2}\right)^n\right| \to \infty$
I tried proving this with SML by myself. Even with SML, it is difficult.
Oct
6
comment Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $
@Leucippus That one is easy to see when you notice that the integral with $\cos \pi x$ vanishes. My solution came out like this when started computing that one instead.
Oct
6
comment Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $
@user153012 I do not know for sure, but the calculation would be much more difficult for other limits.
Oct
5
comment Surprising identities / equations
@Potato I posted a solution to this one on math.stackexchange.com/questions/958624/….
Oct
5
revised Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $
edited body
Oct
5
answered Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $