Given that $(a_n)_{n\to\infty}$ satisfies the condition $\vert a_n-a_{n+1}\vert <\frac{1}{4^n}$ $\forall n\in\mathbb N$ prove that $\lim_{n\to \infty} a_n={}$finite
I wanted to prove that using Cauchy's convergence test. I need to show that $\vert a_{m}-a_{n}\vert <\epsilon\quad\forall m,n\in\mathbb N$. $\quad$ So that's what I did: Let assume without loss of generallity that $m>n$ than
\begin{align} & \vert a_m-a_n\vert=\vert a_m-a_{m-1}+a_{m-1}+\cdots-a_{n+1}+a_{n+1}-a_n\vert \\[10pt] \le {} & \vert a_m-a_{m-1}\vert + \vert a_{m-1}-a_{m-2}\vert+\cdots+\vert a_{n+1}-a_n\vert \\[10pt] < {} &\frac{1}{4^m}+\frac{1}{4^{m-1}}+\cdots+\frac{1}{4^{n+1}} <\frac{1}{4^n}+\frac{1}{4^n}+\frac{1}{4^n}+\cdots+\frac{1}{4^n}<\frac{n}{n^2}=\frac{1}{n}<\epsilon \end{align}
So I proved it eusing Cauchy's convergence test but Im not sure that I can say for sure that $\frac{1}{4^n}+\frac{1}{4^n}+\frac{1}{4^n}+\cdots+\frac{1}{4^n}<\frac{n}{n^2}$ because when I said that I assumed that there is only $n$ terms in the left equation, but what if there is more? I can't know for sure, or maybe because $n\to \infty$ it donsn't relly matter so what I'm asking basically that if my proof holds?