Limit of the geometric sequence $\{r^n\}$, with $|r| < 1$, is $0$? Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.


I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(
Let $a_n=|r|^n$ for $n\ge1$, 
so $(a_n)$ is a decreasing sequence which is bounded below by zero 

and therefore converges, so let $\displaystyle\lim_{n\to\infty}a_n=L$.

Since $a_{n+1}=|r|a_n$, and $(a_{n+1})$ is a subsequence of $(a_n)$, by the theorem that all subsequence of a convergent sequence converge to the same limit$,  \;\;L=|r|L\implies L=0\;\;$ (for $r\ne0$).

Since $\displaystyle\lim_{n\to\infty}|r|^n=0$,  $\displaystyle\lim_{n\to\infty}r^n=0$ also.
This is an answer I had earlier posted.

This is not a rigorous proof but helps in visualisation.

If $$-1&lt;x&lt;1$$ and you raise it to any power $n &gt; 1 $ the value of it decreases.

Eg say$ x = 0.1 , x ^2  = 0.01, x^3 = 0.001$ or $x ^n = 0.1 ^ n = \frac{1}{10^n}$ and as $ n$ keeps getting bigger , the denominator approaches $\infty $ and the numerator approaches $0 $