Proving that length of closed intervals in fat Cantor Set approaches 0

Question

I am looking to show that $l_n$ where $l_n$ denotes the length of closed subintervals in the nth stage of the fat cantor set approaches 0. I define the fat cantor set $C_\alpha$ to be the set constructed by first deleting the intervals $(1/2 - \alpha/4, 1/2+\alpha/4)$ from $[0,1]$, then deleting intervals centered at the midpoint of the new closed intervals, and taking off $\alpha/8$ from each side, and so on and so forth. To show these lengths go to 0, I first outlined a recursive formula for $l_n$ given by: $$ l_n = \frac{1}{2}\left( l_{n-1}-\frac{\alpha}{2^{n+1}}\right) $$ It was easy to show the sequence was monotone decreasing, but now I must show that it has infimum 0. I am not entirely sure how to do this, as this recursive formula is annoying to work with. I was wonderign if anyone could lead me down the right path. Please note that this is a homework problem, so a hint would be preferable to a full answer. Thank you.

Answer 1

Hint: Just ignore the complicated term with $\alpha$, and write your recurrence as $l_n\leq \frac{1}{2} l_{n-1}$. What upper bound on $l_n$ does this give by induction?

More generally, when you want to prove things about limits, you just want inequalities. So you should not worry about getting some exact formula you can work with; instead you just want simple estimates which you can prove are upper and lower bounds and are good enough for the final conclusion you want.