I know that this is a cauchy sequence. But I am interested in what I wrote in line 8 to 12. I was wondering if it was correct at all? I am trying to learn the bounded sequences proof
We assume that the terms of a sequence ${S_n}$ satisfy this property $\forall n$
$$|a_{n+1}-a_n| \leq \Big| \frac{1}{2^n} \Big| $$
Now I would like to show that $\{a_n\}$ converges. Specifically towards $0$
As $ \ n \longrightarrow \infty \ \text{, we have } \dfrac{1}{2^n} \longrightarrow 0$
Also $|a_{n+1}-a_n| \geq 0$
So we have, as $ \ n \longrightarrow \infty$
$$\text{(line 8) }\hspace{1cm} 0 \leq |a_{n+1}-a_n| \leq 0$$
$$\implies \hspace{1.2cm} \ 0 \leq |a_n - a_{n+1}| \leq 0$$ by triangle ineq.
$$\implies \hspace{1.2cm} 0 \leq |a_n| - |a_{n+1}| \leq 0 $$
$$\text{(line 12) } \implies \hspace{0.2cm} \ 0 \hspace{.2cm} \leq \hspace{.2cm}|a_n| \hspace{.2cm} \leq \hspace{.2cm}0$$
Can I use this to say that this sequence is bounded and thus converge to $0$
Note: To be honest, I feel this is correct but I am not satisfied with the argument.
Hint
Set $S_n=\sum_{k=0}^n\frac{1}{2^k}$. It's a convergent sequence. Moreover, $$|a_n-a_m|\leq |S_n-S_m|.$$
I let you conclude.
Notice that there is no reason that $(a_n)$ converges to $0$.