<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] To prove that the series satisfies the Cauchy condition you have to estimate, for $n,p\in\mathbb{N}$, the difference
$$
\sum_{j=n}^{n+p} \frac{1}{2^j}  = \frac{1}{2^{n-1}}\left(1 - \frac{1}{2^{p+1}}\right) < \frac{1}{2^{n-1}}\,.
$$
Given $\varepsilon > 0$, it is enough to choose $N > 1 - \log_2\varepsilon$ so that
$$
\sum_{j=n}^{n+p} \frac{1}{2^j} < \varepsilon
\qquad
\forall n> N, \ p\in\mathbb{N}. 
$$
</s>