How to proof that this sequence $\frac{a^n}{1-a}$ converges into zero

Question

Also, I need to denote that $a$ lies in this interval $(0,\,1)$. So I have tried with different convergent sequences but any of them did the trick to delimit the original sequence, any ideas of how can I find a sequence that could let me set an $\varepsilon$ in which I can found an $N$ for all $n \gt N$, where the distance of the elements of the sequence are less than an $\varepsilon \gt 0$?

Answer 1

If $a\in(0,1)$ and $\varepsilon>0$, then by monotonicity of the logarithm, we have for any positive integer $N$: $$a^N < \varepsilon \iff \log(a^N)<\log\varepsilon \iff N > \frac{\log \varepsilon}{\log a}. $$ (Note the switch in sign due to $\log a < 0$.) Given such an $N$, we have $$a^n < a^N <\varepsilon,\quad n\geqslant N,$$ as the sequence $a^n$ is strictly decreasing.

(To deal with the constant factor of $\frac1{1-a}$ we need only vary the formula for $N$ slightly...)

Answer 2

Using Ratio test,

$$\lim_{n\to \infty }\frac{\frac{a^{n+1}}{1-a}}{\frac{a^n}{1-a}}=a\in (-1,1).$$ Therefore, the sequence converge to $0$.