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As a part of the proof of recurrence criterion, (Page 22 here: http://web.math.ku.dk/noter/filer/stoknoter.pdf), it is shown that $(P^n)_{i,j} = \sum_{m=1}^n (P^{n-m})_{j,j}f_{ij}^{(m)}$. This makes sense. The next step is to sum over $n = 1,\dots,N$. Therefore,

$$\sum_{n=1}^N (P^n)_{i,j} = \sum_{n=1}^N \sum_{m=1}^n (P^{n-m})_{j,j} f_{ij}^{(m)}$$

However, this directly leads into an equality

$$ = \sum_{m=1}^N \sum_{n=m}^N (P^{n-m})_{j,j} f_{ij}^{(m)}$$

This is likely a general summation question, but I cannot see what principles are at play to reindex as such.

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  • $\begingroup$ Confined to the unit square $[0,1]^2$, the region of integration defined by $y \geq x$ is $\int_0^1 \int_x^1 dy dx = \int_0^1 \int_0^y dx dy$. Here, $1$ plays the role of $N$, the $x$ limits play the role of $m$, and the $y$ limits play the role of $n$. If that doesn't help then replace $(P^{n-m})_{j,j} f_{i,j}^{(m)}$ with $a_{n, m}$ and write it out for $N = 3$. Arrange the terms into a triangle, as in the region of integration. $\endgroup$ – snar May 15 at 19:23

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