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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).

Frank
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    Your $s$ is the number of terms, so differentiating with respect to that sounds very fishy to me. I would try using the sum formula for the geometric sums. – Jyrki Lahtonen Mar 29 '13 at 16:03
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    If you're sold on getting L'Hospital in there, the denominator is $(1-(ab)^{s+1})/(1-ab)$, so its correct derivative is $-\ln(ab)(ab)^{s+1}/(1-ab)$. Do the same with the numerator. – Jyrki Lahtonen Mar 29 '13 at 16:07

4 Answers4

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Hint: Use the closed form expression $$1+r+r^2+\cdots +r^{n}=\frac{1-r^{n+1}}{1-r}.$$ Note that this only applies for $r\ne 1$.

André Nicolas
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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 $$\lim_{s \to \infty} \frac{(ab)^{s+1}-ab}{(ab)^{s+1}-1}=\lim_{s \to \infty} \frac{1-\frac1{(ab)^s}}{1-\frac1{(ab)^{s+1}}}=1$$

lab bhattacharjee
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Edit: Since you've added in the assumption that $a,b\in[0,1]$, then $0\le ab\le 1$, and so the numerator and denominator only diverge in the case that $a=b=1$. In that case, L'Hopital's rule does indeed yield a limit of $1$...which is precisely $ab$.

Otherwise, we have $$1+ab+\cdots+(ab)^s=\frac{1-(ab)^{s+1}}{1-ab},$$ whence $$\frac{ab+\cdots+(ab)^s}{1+ab+\cdots+(ab)^s}=1-\frac1{1+ab+\cdots+(ab)^s}=1-\frac{1-ab}{1-(ab)^{s+1}},$$ so since $0\le ab<1$, then $(ab)^{s+1}\to 0$ as $s\to\infty$, so again we have $ab$ as the limit.

Cameron Buie
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  • Sorry, I forgot to mention that. I know it's 1, but according to the solution of the tutorial I'm working through, the answer is $ab$, hence my confusion. – Frank Mar 29 '13 at 16:01
  • It can be $1$, but it can also be $ab$. There may have been additional assumptions about $ab$ that you didn't mention. – Cameron Buie Mar 29 '13 at 16:19
  • I added those assumptions in my edit. Those are definitely relevant assumptions (as I see now) and I just forgot to include them. – Frank Mar 29 '13 at 16:28
  • @CameronBuie, $ab$ can be $=1$ right? – lab bhattacharjee Mar 29 '13 at 16:38
  • @labbhattacharjee: It can, but I dealt with the case $ab=1$ separately from the case $0\le ab<1$. Moreover, if we were to allow $|ab|>1$, then the limit would be $1\neq ab$. – Cameron Buie Mar 29 '13 at 16:40
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The derivative (with respect to $x$) of $a^x$ is not $x a^{x-1}$ but $ln(a)a^x$, since $a^x=e^{x ln(a)}$.

You can solve your problem by noticing that $\displaystyle \frac{ab + (ab)^2 + ... (ab)^s}{1 +ab + (ab)^2 + ... (ab)^s}=1- \frac{1}{1 +ab + (ab)^2 + ... (ab)^s}$.

Amzoti
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Taladris
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