Sigma notation

Sigma notation writes a sum of many terms as one expression, with the Greek capital letter $\Sigma$. The index $k$ starts at the lower bound $m$ written under the $\Sigma$ and goes up one at a time to the upper bound $n$ written above it. Each value it takes gives a term $a_k$, and the terms, not the values of the index, are added:

When $m \le n$ there are $n - m + 1$ terms: both bounds are included, unlike range(m, n) in some programming languages, which stops before $n$. The bounds can be any integers, and the same sum is also written $\sum_{m \le k \le n} a_k$.

Here each term is a bar standing over its index, and beside the bars the running-total diagram traces the successive totals, one arrow per term. You can choose the bounds and the term, look at any one term to see which index gave it and what the total is up to it, or add the terms one at a time and keep a row of the working for each.

The index means something only inside the sum. Rename it everywhere, under the $\Sigma$ and in the term, and nothing changes: $\sum_{k=1}^{3} k^2 = \sum_{j=1}^{3} j^2 = 1 + 4 + 9 = 14$. That is why it is called a dummy variable. Renaming it in only one place changes the expression: in $\sum_{j=1}^{3} k^2$ the term does not depend on $j$, so the sum is $3k^2$. The index's job is to say which terms there are, so it need not appear in the term at all: for $m \le n$, a constant summed from $k = m$ to $n$ is that constant added $n - m + 1$ times, $\sum_{k=m}^{n} c = (n - m + 1)\,c$.

When there are terms ($m \le n$), moving a bound by one changes the sum by exactly one term at that end, because both ends are included: raising the upper bound brings in the next term, and raising the lower bound drops the first.

If the upper bound is below the lower one, as in $\sum_{k=5}^{4} a_k$, the index takes no value and there are no terms. By the usual convention this empty sum is 0: a total starts at 0, and nothing is added to it. The convention keeps the step from one bound to the next true at the very start; $\sum_{k=1}^{n} a_k = \sum_{k=1}^{n-1} a_k + a_n$ holds at $n = 1$, where the sum on the right has no terms. A sum of no terms is not the same as terms that cancel: in $\sum_{k=1}^{6} (-1)^k = 0$ all six terms are there.