Is $\sum _{t=1}^m \sum _{k=t}^m (a_{k+1}-a_k)b_t$ equal to $\sum _{k=1}^m \sum _{t=1}^k (a_{k+1}-a_k)b_t$?
If not, then how to derive equations for temporal difference algorithm in machine learning: page 14 at the end, and page 15 equation (3).
http://webdocs.cs.ualberta.ca/~sutton/papers/sutton-88.pdf
It seems like an error?
$$\begin{align} \sum_{t=1}^m\sum_{k=t}^m(a_{k+1}-a_k)b_t & =\sum_{t=1}^m((a_{t+1}-a_t)b_t+(a_{t+2}-a_{t+1})b_t+\cdots+(a_{m+1}-a_m))b_t\\ & =(a_2-a_1)b_1+(a_3-a_2)(b_1+b_2)+\cdots+(a_{m+1}-a_m)(b_1+\cdots+b_m)\\ & = \sum_{k=1}^m\sum_{t=1}^k(a_{k+1}-a_k)b_t. \end{align}$$