Cauchy sequence $a_n = 1 + \frac12 + \frac14 + ... + \frac{1}{2^n}$ For the sequence $a_n = 1 + \frac12 + \frac14 +  ... + \frac{1}{2^n}$, $n \ge 1$, find a formula $N = N(\epsilon)$ such that
for all $\epsilon > 0$ and for all $m,n \ge N(\epsilon)$, $|a_m - a_n| < \epsilon$.
I have tried many things but it's just not working. I know that the condition is the definition of Cauchy sequences. But that didn't help.

Hint:
\begin{align*}
a_n
&= 1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} \\
&= \frac{1}{2^n} (2^n + 2^{n-1} + 2^{n-2} + \cdots + 1) \\
&= \frac{2^{n+1} - 1}{2^n} = 2 - \frac{1}{2^n}
\end{align*}
so
\begin{align*}
|a_n - a_m| &= \left| \left( 2 - \frac{1}{2^n} \right) - \left( 2 - \frac{1}{2^m} \right) \right| \\
&= \left| \frac{1}{2^n} - \frac{1}{2^m} \right| \\
&< \frac{1}{2^m} + \frac{1}{2^n} \\
&\le \frac{1}{2^N} + \frac{1}{2^N} = \frac{2}{2^N}
\end{align*}
Pick $N(\epsilon)$ so that $\frac{2}{2^N} < \epsilon$.
