Double Summation Switch

For these things, I often find it helpful to think of the direction of the summation.

Think of it as a matrix, where $i$ denotes the row and $j$ the column. The first sum says that: in the first row, when $i=1$, we sum the entries in columns 1 to infinity. That is, we sum the entire (infinite...) row starting at 1. Then, we move to the second row, such that $i=2$, and sum the entire row from 2 and upwards. This pattern continues - in row $i$, we sum all entries from column number $i$ and to the right of it. Thus, we are doing horizontal summation.

It is, however, possible to do the summation vertically as well. To sum the exact same entries but in this way instead, we start with column $j=1$. Here, we want the entry in the first row only. Then, we move to $j=2$ and sum elements 1 and 2. When $j=3$, we sum elements 1, 2, and 3 since we want everything to the right of, and including, the main diagonal. Thus, the pattern that we find is that $i\leq j$. Hence, having $j$ as the outer sum we let it go to infinity to get the entire upper triangular part of the matrix and let the sum for $i$ stop at $j$.

Here's a picture too:

Instead of directly switching the sums, you can use an intermediate step with only one sum, for example:

$$\sum_{i=1}^{n}\sum_{j=1}^{i}f\left(\left(i,j\right)\right)\\ = \sum_{t\in\left\{ \left(i,j\right)\mid j\le i\le n\right\} }f\left(t\right)\\ = \sum_{j=1}^{n}\sum_{i=j}^{n}f\left(\left(i,j\right)\right)$$

This has the advantage that it works with pretty much arbitrary amounts of nested summations and it's also easier when you don't want to draw the matrix suggested by @hejseb.