How to calculate $ \lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s} $ I'm trying to calculate this limit expression:
$$ \lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s} $$
Both the numerator and denominator should converge, since $0 \leq a, b \leq 1$, but I don't know if that helps. My guess would be to use L'Hopital's rule and take the derivative with respect to $s$, which gives me:
$$ \lim_{s \to \infty} \frac{s (ab)^{s-1}}{s (ab)^{s-1}} $$
but this still gives me the non-expression $\frac{\infty}{\infty}$ as the solution, and applying L'Hopital's rule repeatedly doesn't change that. My second guess would be to divide by some multiple of $ab$ and therefore simplify the expression, but I'm not sure how that would help, if at all. 
Furthermore, the solution in the tutorial I'm working through is listed as $ab$, but if I evaluate the expression that results from L'Hopital's rule, I get $1$ (obviously). 

If $ab=1,$
$$ \lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s}= \lim_{s \to \infty} \frac{s}{s+1}=\lim_{s \to \infty} \frac1{1+\frac1s}=1$$
If $ab\ne1, $
$$\lim_{s \to \infty} \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s}$$
$$=\lim_{s \to \infty} \frac{(ab)^{s+1}-ab}{(ab)^{s+1}-1}$$
If $|ab|<1, \lim_{s \to \infty}(ab)^s=0$ then $$\lim_{s \to \infty} \frac{(ab)^{s+1}-ab}{(ab)^{s+1}-1}=ab$$
Similarly if $|ab|>1,\lim_{s \to \infty}\frac1{(ab)^s}=0$
then $$\lim_{s \to \infty} \frac{(ab)^{s+1}-ab}{(ab)^{s+1}-1}=\lim_{s \to \infty} \frac{1-\frac1{(ab)^s}}{1-\frac1{(ab)^{s+1}}}=1$$
