I'm trying to simplify this summation $\sum _{k=1}^i 2^{h-k}$ where $1\leq i\leq h$. After performing variable substitution, however, the upper bound becomes less than the lower bound:
\begin{align} j=h-k\\ k=1\Rightarrow j=h-1\\ k=i\Rightarrow j=h-i\\ \sum _{k=1}^i 2^{h-k} &= \sum_{j=h-1}^{h-i} 2^j\\ &=\sum _{j=h-i}^{h-1} 2^j \text{ ................. bounds swapped}\\ &=\sum _{j=0}^{h-1} 2^j-\sum _{j=0}^{h-i-1} 2^j\\ &=\frac{2^h-1}{2-1}-\frac{2^{h-i}-1}{2-1}\\ &=2^h-2^{h-i} \end{align}
Would it be legal to switch the bounds in this case?
For me, it is rather the first step that looks illegal, i.e., I would allow only the direct transition $$\tag1\sum_{k=1}^i 2^{h-i}=\sum_{j=h-i}^{h-1}2^j,$$ but that depends on how (if at all) you defined $\sum$ when the top bound is below the bottom bound. For example, I would prefer the equation $$\tag2 \sum_{k=n}^m x_k=x_m+\sum_{k=n}^{m-1}x_k = x_n+\sum_{k=n+1}^m x_k$$ to always be true (and therefore $\sum_{k=n}^m=-\sum_{k=m+1}^{n-1}$) instead of $$\tag3 \sum_{k=n}^mx_k=\sum_{k=m}^nx_k,$$ (you can't have both) but your mileage may vary.
But I think the most common definition with respect to wrongly ordered bounds is that $$ \sum_{k=n}^mx_k=0\quad\text{if }m<n$$ (and again, this means you must use the transition $(1)$ directly). This comes naturally (and thereby differs from $(2)$) by interpreting the index bound notation as a shorthand for summing over a finite set: $$ \sum_{k=n}^mx_k:=\sum_{k\in\{k\in \Bbb Z\mid n\le k\le m\}}x_k.$$