Why is this equality true? I'm working with a proof of local and global truncation errors for the euler method in numerical methods, and I can't seem to understand the equality and the less/equal seen below. At first I thought it had to do with geometrical sums, but I still don't understand it, any help would be appreciated:
$$1+e^{Lh}+e^{2Lh}+...+e^{(i-1)Lh} = \frac{e^{iLh}-1}{e^{Lh}-1} \le \frac{e^{L(t_i-a)}-1}{Lh}$$

The equality is indeed the sum of the geometric series (although $i$ is somewhat confusing: it is the number of intervals, not $\sqrt{-1}$ :-) ).
The inequality arises from two things: $iLh = L(t_i - a)$, so the numerators are equal. For the denominators, $Lh$ is apparently assumed small, so you can approximate $e^{Lh}$ by the first two terms of its Taylor expansion. Since the rest of the expansion is positive, leaving it out reduces the total:
$$ e^{Lh} -1 \ge Lh$$
so inverting inverts the inequality.
EDIT: The inequality is true even if $Lh$ is not small, but when it is small, the inequality is "good" in the sense that it provides a tighter bound.
