Background:Suppose that we are given the following problem:
We are given $2n$ points $A_1,A_2,...,A_n$ and $B_1,B_2,...,B_n$ on a circle $\mathcal{C}$.Prove that there is a point $P\in\mathcal{C}$ such that $\sum_{i=1}^{n}PA_i=\sum_{i=1}^{n}PB_i$.
I'm not interested in a solution(I know a solution),but I just want to know,if this approach is legit:
Maybe a solution: For a point $P\in\mathcal{C}$ define $f(P)=\sum_{i=1}^{n}PA_i-\sum_{i=1}^{n}PB_i.$ Note that $$\sum_{P\in\mathcal{C}}f(P)=\sum_{P\in\mathcal{C}}\sum_{i=1}^{n}PA_i-\sum_{P\in\mathcal{C}}\sum_{i=1}^{n}PB_i=\sum_{i=1}^{n}\sum_{P\in\mathcal{C}}PA_i-\sum_{i=1}^{n}\sum_{P\in\mathcal{C}}PB_i.$$
Now,for a fixed point $X\in\mathcal{C}$ the sum $\sum_{P\in\mathcal{C}}PX$ is the same,so $\sum_{P\in\mathcal{C}}f(P)$ must be $0$.Hence the exist points $P_1,P_2\in\mathcal{C}$ such that $f(P_1)\le 0\le f(P_2)$ and by continuity of $f(P)$ it follows that there must be some point $Q$ on the arc $P_1P_2$ such that $f(Q)=0$.
Main Question: The above solution is like a cheat,since the sum $\sum_{P\in\mathcal{C}}PX$ is infinite and we can't really subtract two infinite sums.But the sums,even if they are infinite,they seem to be the "same".
My questions are:Can we play with infinite sums like in the above solution?Is really $f(P)=0$ or is just a sum that we can't evaluate?And how can we really deal with infinite sums?For example,it would seem logic that $\sum_{-1}^{1}x=0$,since we can pair the numbers $(x,-x)$,but then we end up with the infinite sum $\sum_{x=0}^{1}(x-x)$,witch seems to be $0$,but is is really $0$?
By the way,the solution I know to the problem uses integrals,so is integration a way to deal with infinite sums?
Well, the first problem is the circle is an uncountable set, so the usual way of defining infinite sums with sequences and partial sums doesn't really work. There's a way to extend sums to uncountable sets, but the sum will diverge unless all but a countable subset of the terms are zero. There are also alternative summation methods which are interesting and can sometimes give finite sums for series that are normally considered divergent, but I don't know if they'd help here.
Beyond that, it's true that $\sum a_n + \sum b_n = \sum (a_n+b_n)$ if all three are defined, but the problem you have is rearrangement. Jumbling the terms of a series changes its sum in general. To get the cancellation in your argument you're mixing up the order of your series, so you would have to justify that by proving it didn't change the sum.