<s>[INST] Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n $ is Cauchy Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n$ is cauchy
I know from the definition of Cauchy that |$x_n$-$x_m$|< ϵ but how do you do this with |$\frac{1}{2^n}- \frac{1}{2^m}$|
what I've tried:
if $n\gt m$ then
$$ |\frac{1}{2^n}- \frac{1}{2^m}| \le |\frac{2^m-2^n}{2^n2^m}| \le |\frac{2^m +2^n}{2^{n+m}}| \le \frac{2^n+2^n}{2^{2n}}= \frac{1}{2^{n-1}}     \le \frac{1}{2^{N-1}} \le \epsilon   $$
and rearrange to get N $ \ge 1+ \frac{ln(\epsilon)}{ln(2)}$
is this correct?
 [/INST] Let $S_n=\sum_{k=1}^n\frac{1}{2^k}$.  Then, we have for any $\epsilon>0$
$$\begin{align}
\left|S_n-S_m\right|&=\left|\sum_{k=\min(m,n)+1}^{\max(m.n)}\frac{1}{2^k}\right|\\\\
&=\left|\frac{1}{2^{\min(n,m)}}-\frac{1}{2^{\max(m,n)}}\right|\\\\
&\le \frac{1}{2^{\min(n,m)-1}}\\\\
&<\epsilon
\end{align}$$
whenever $\min(n,m)>1-\frac{\log(\epsilon)}{\log(2)}$
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