For which value of "d" cantor set has lebesgue measure zero?

Question

I tried to generalise the cantor set. I retain "$d<1$" size interval at both end and removed "1-2d" size interval, and process is same further. I have curious about what value of "d" this cantor set has Lebesgue measure zero? I tried upto this: in first stage we removed interval with size "1-2d", in second stage we removed two copies of "$ 1-2d^2 $" size intervals, in third stage we removed four copies of "$ 1-2d^3$" size of intervals, and so on, at $n^{th}$ stage we removed $2^{n-1}$ copies of "$1-2d^n$" size intervals. So I came upto following: For which value of "d" following equation hold? $$ \sum_{n=1}^{\infty} 2^{n-1} (1-2d^n) =1. $$ I solve this by ratio test and I get answer d=1/2. I guess $0<d<1/2$ will satisfy this. Am I going right direction? Please help?

Answer 1

It seems to me that at the first step you have two lengths of size $d$ so you have removes a length of size $1-2d$ while After the second step you have four steps of size $d^2$ and so have removed two lengths each of size $d(1-2d)$ at that step

At the $n$th step you remove $2^{n-1}$ lengths of size $d^{n-1}(1-2d)$ and are left with $2^n$ lengths of size $d^n$, so in total you remove $$\sum_{n=1}^{\infty} (2d)^{n-1} (1-2d)$$ which is $1$ for $0\lt d \lt \frac12$