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 formula $$\sum^i_{n=0} x^n = \frac{x^{i+1}-1}{x-1}$$ holds for any $x\ne 1$ [this is easily seen for $i=1$ and then proven by induction]. Just plug in $x=e^{Lh}$. The inequality follows since $e^{Lh} -1 \ge Lh$ and presumably $ih = t_i -a$.
