@robjohn @robjohn do you remember any sum involving harmonic number that yields $\displaystyle \frac{3\zeta(3)}{16}$? The idea is tha I have the feeling I met that result somewhere but I don't recollect where exactly.
Hey guys--I'm not sure if this is the right place to ask, but I have a graph theory question I am having an impossible time finding an answer to. Does anyone know graph theory well? And please forgive me if this is the wrong place to ask.
Ah well. I am trying to show that the graph of any convex polyhedron is 3-connected. I am pretty decent at graph theory but I have almost no feel for how a convex polyhedron behaves and I am just not sure how to proceed
@Mike @KarlKronenfeld This is how I did it. Let $A$ be commutative, and $\mathfrak a+\mathfrak b=A$. Say $a'+b'=1$, and $x\in \mathfrak a\cap\mathfrak b$, $x=a+b$. Then note $a,b\in\mathfrak a\cap\mathfrak b$, and thus $x=1\cdot a+1\cdot b=a'a+b'a+a'b+b'b$ is in the product ideal.
@Chris'ssis I have a lot of results that deal with $\zeta(3)$ such as this one, which says that $$\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)$$ I have not found any with that particular rational coefficient yet.
@r9m If a rectangular board can be filled up by $1\times m$ and $n \times 1$ strips, then prove that this could be done with only one of the type only.
@Chris'ssis a few days ago you showed a really interesting series .. $\sum \sin(\pi\sqrt{n^2+1})$ .. does it converge to anything interesting ? I mean is the point of convergence known ?
@Mike: ... that was what I had in mind but my programmers mentality, man. It's really bugging me out. Don't I have to first declare the matrix before I say that? Can't I do both at once.?
Something like $$A = \text{Matrix}(a_{ij} = f(i,j)) $$
@robjohn I want to show you a proof that contains that result, but I bit later. I 'm working on it right now and answer a lot of phone calls (unfortunately).
