Summation/Sigma notation There are lots of variants in the notation for summation. For example, $$\sum_{k=1}^{n} f(k), \qquad \sum_{p \text{ prime}} \frac{1}{p}, \qquad \sum_{\sigma \in S_n} (\operatorname{sgn} \sigma) a_{1 , \sigma(1)} \ldots a_{n , \sigma(n)}, \qquad \sum_{d \mid n} \mu(d).$$ 
What exactly is a summation? How do we define it? Is there a notation that generalizes all of the above, so that each  of the above summations is a variant of the general notation? Are there any books that discuss this matter? 
It seems that summation is a pretty self-evident concept, and I have yet to find a discussion of it in a textbook. 
 A: Except for the case of the upper and lower limit, all the other summations are really just sums of the form $$\sum_{P(i)} f(i)$$
Where $P$ is a unary predicate in the "language of mathematics", and $f(i)$ is some function which returns a value that we can sum. In the case of the sum of prime reciprocals $P(i)$ states that $i$ is a prime number and $f(i)=\frac1i$. In the second sum, $P(i)$ was $i\in S_n$, and $f(i)$ was that summand term. And so on. 
A: The wikipedia explains this (I think) pretty well, give it a try:
http://en.wikipedia.org/wiki/Summation#Notation
A: The OP stated

It seems that summation is a pretty self-evident concept

The starting ground, when adding up only a finite number of terms, is to understand two concepts:


*

*Addition is associative

*Addition is commutative


Allowing the sigma sum notation to 'process' an infinite number of terms, the serious student will be have to learn about absolute convergence.
Here we will provide the theory showing why the sigma notation works (is well-defined) for the finite case (where it is also self-evident to the mathematically inclined).
We use the notation $\bar n = \{1,2,\dots,n\}$. Recall that any permutation $\bar n$ of can be expressed as a product of adjacent transpositions (see wikipedia);
Let $J$ be a finite set with $n$ elements and $f:J \to \mathbb R$ a function 'selecting the summands'. Let $\rho: \bar n \to J$ be a bijection.
We can recursively define a function $F_{(f,\rho)}:\bar n \to \mathbb R$ by
$\tag 1 F_{(f,\rho)}(1) = f(\rho(1))$
$\quad \quad\quad\quad\quad\quad\quad\quad\quad F_{(f,\rho)}(k+1) = F_{(f,\rho)}(k) + f(\rho(k+1))$
It is not difficult to argue that if $\tau$ is an adjacent transposition then 
$\tag 2 F_{(f,\tau \circ \rho)}(n) =  F_{(f,\rho)}(n)$
But then for any permutation $\phi$
$\tag 3 F_{(f,\phi)}(n) = F_{(f,\rho)}(n)$
The following result has been proven to be true.
Theorem: For any two bijections $\rho: \bar n \to J $ and $\phi: \bar n \to J $  we have
$\quad \quad \quad \quad \quad F_{(f,\rho)}(n) = F_{(f,\phi)}(n)$
So we have a fabulous notation, $\sum_{j \in J}  f(j)$. Usually $f$ will be defined on $\bar n$ and you'll be using (letting $a_k = f(k)$)
$\tag 4 \sum_{k=1}^n a_k$
