Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n $ is Cauchy

Question

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?

Answer 1

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 $\min(n,m)>1-\frac{\log(\epsilon)}{\log(2)}$

Answer 2

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 $$ \sum_{j=n}^{n+p} \frac{1}{2^j} < \varepsilon \qquad \forall n> N, \ p\in\mathbb{N}. $$

Answer 3

$S_m,S_n $ are the partial sums in the problem:

$|S_n - S_m| =|1/2^n -1/2^m|;$

Let $\epsilon \gt 0$ be given .

Choose $N_0$ such that $N_0 \gt 2 (1/\epsilon)$. (Archimedes)

Then:

For $m,n \gt N_0 :$

$|S_n-S_m| = |1/2^n-1/2^m| \lt $

$|1/2^n|+|1/2^m| \lt 1/n +1/m \lt$

$2/N_0 \lt \epsilon.$