interchange double summation I am confused I am looking at my professor's note but I don't know how he changes the following double summation. Can someone explain it to me?
$$
\sum_{k = 0}^{\infty}\,\,
\sum_{x = k + 1}^{\infty}\operatorname{f}\left(x\right) = \sum_{x = 1}^{\infty}\,
\sum_{k = 0}^{x - 1}
\operatorname{f}\left(x\right)
$$
 A: There are obvious convergence issues to be worried about with such an interchange. But assuming everything converges uniformly, we can make the interchange as follows (let me change $x$ to $n$ since $x$ will be a little awkward in the discussion that follows). This is hard to describe but easy to picture. If you get confused, my suggestion is to draw everything out.
The value of
$$\sum_{k=0}^\infty \sum_{n=k+1}^\infty f(n)$$
is geometrically simple. Think of the pairs $(k,n)$ as integer points in $\mathbb{R}^2$, say. Then the sum for $k=0$ corresponds to the sum over all integer points with $x$-coordinate $0$ and $y$-coordinate greater than or equal to $1$, which is an infinite column of points. Likewise, the sum for $k=1$ corresponds to the sum over all integer points with $x$-coordinate $1$ and $y$-coordinate greater than or equal to $2$, another infinite column. In general, the sum will be over all integer points with non-negative $x$-coordinate and $y$-coordinate satisfying $n \ge k+1$. This is a triangular region (draw it out), and the first sum corresponds to summing over the triangle column by column.
Of course, another equivalent way of summing over this triangle is not column by column, but row by row. In this case, the first row is at $y=1$, and consists of only the point with $x=0$. The second row is at $y=2$, and consists of the points with $x=0$ or $x=1$. In general, the $n$th row has $y$-coordinate $y=n$ and $n$ total points: $x=0,1,\cdots ,n-1$. Summing this triangle row by row gives us
$$\sum_{n=1}^\infty \sum_{k=0}^{n-1}f(n).$$
The two sums are equal since they sum over the exact same set of points, just in two different orderings.
