The definition of a Cauchy sequence I'm considering is $\forall \epsilon>0, \ \exists N(\epsilon)\in \mathbb{N}; \ \forall n,m > N(\epsilon) \ |a_n-a_m| < \epsilon$. I think my proof is fine up to a certain point, but then I have some trouble articulating my thoughts. Help there would be much appreciated. Also I'll be using the fact that $\Sigma_{k=1}^{n} q^k = \frac{1-q^{n+1}}{1-q}$ in this proof.
Let $s_n = \Sigma_{k=0}^{n} \frac{1}{2^k}$. Then we have $|s_n-s_m|<\epsilon \impliedby |\Sigma_{k=1}^{n} \frac{1}{2^k} - \Sigma_{k=1}^{m} \frac{1}{2^k}|<\epsilon \impliedby 2|(1/2)^{m+1}-(1/2)^{n+1}|<\epsilon$.
Here is where I think things fall apart a little. If we let $a_n = (1/2)^n$, we know that $a_n \rightarrow 0$, so we know that $\forall \epsilon_1 >0, \ \exists N_1(\epsilon_1)\in \mathbb{N}; \ \forall n>N_1(\epsilon_1) \ |(1/2)^n-0|<\epsilon_1$. If we let n,m > $N_1(\epsilon_1)$ then we have $2|(1/2)^{m+1}-(1/2)^{n+1}| < 2|(1/2)^{m+1}|+2|(1/2)^{n+1}| < 2(\epsilon_1)+2(\epsilon_1) < \epsilon$. So should we let $\epsilon_1 < \epsilon/4$ then we have our desired conclusion that $|s_n-s_m|<\epsilon$ and so the sequence is Cauchy.
Any and all input is appreciated. Thanks!
First, note that if $n \ge m \ge N$, then $|s_n-s_m| = \sum_{k=m}^{n-1} {1 \over 2^k} \le \sum_{k=N}^{\infty} {1 \over 2^k} = 2{1 \over 2^N}$.
Now choose $\epsilon>0$ and choose $N$ such that $2{1 \over 2^N} <\epsilon$.