I have the following interval: $$I_n = \Big(a_n - \frac{\epsilon}{2^{n+1}}, a_n + \frac{\epsilon}{2^{n+1}}\Big)$$ I have then proceeded to find the Lebesgue measure as: $$\sum_{n=1}^{\infty}m(I_n) = \sum_{n=1}^{\infty}\Bigg(a_n+\frac{\epsilon}{2^{n+1}}\Bigg)-\Bigg(a_n - \frac{\epsilon}{2^{n+1}}\Bigg) = \sum_{n=1}^{\infty}\Bigg(\frac{2 \epsilon}{2^{n+1}}\Bigg) = \epsilon \sum_{n=1}^{\infty}\Big(\frac{1}{2}\Big)^n$$ But how does: $$\epsilon \sum_{n=1}^{\infty}\Big(\frac{1}{2}\Big)^n = \epsilon$$???
Any help would be greatly appreciated.
Because $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = 1$$
Indeed, $$\sum_{n=1}^N \left(\frac{1}{2} \right)^n = \frac{1}{2}. \frac{1- \left( \frac{1}{2}\right)^{N-n}}{1 - \frac{1}{2}} \underset{N\to +\infty}{\longrightarrow} \frac{1}{2} .\frac{1}{1 - \frac{1}{2}} = 1$$