/sci/ - Science & Math

[math] \sum_{n=0}^m x^n = \dfrac{1-x^{n+1}}{1-x} [/math]

so for |x|<1

[math] \lim_{m\to \infty} \sum_{n=0}^m x^n = \dfrac{1}{1-x} [/math]

Now the special case x=1/2 gives

[math] 1 + \lim_{m\to \infty} \sum_{n=1}^m \dfrac{1}{2^n} = \dfrac{1}{1-1/2} [/math]

>>9323458
I try my best. I'm also >>9322704, maybe you have an input.

>>9323427
Let X denote the cateogry with two obect and else nothing more than two parallel arrows that go from one object to the other. A graph is a functor from X to the category of sets.

This definition works because the sets in the object-image of the functor can be understood as the edges and vertices, respectively, and the arrow-image are two functions that select, for each edge, a beginning and an end.