If $0If $0 < r < 1$ show that $r^n$ goes to $0$ as $n \to 1$.
If $|r| < 1$ then
$r^2  < r$
similarly $r^4 < r^3 < r^2 < r$
so $r^n$ as $n \to +\infty$ will be equal to $1$
how can i finish this problem?
 A: You seem to have determined (and be able to prove, if asked!) that the sequence $r^n$ is descending. It is also clearly bounded from below (by zero), so by the theorem on monotonic sequences it converges to some limit $A$. But then the subsequence $r^{n+1}$ alsot tends to $A$ as $n\to\infty$. Therefore
$$
A=\lim_{n\to\infty}r^n=\lim_{n\to\infty}r^{n+1}=r\lim_{n\to\infty}r^n=rA.
$$
From this equation we can solve $A=0.$
A: Verify the definition of limit:
Fixed $\varepsilon >0$, you need to find an $N$ such that $|r^n| = |r|^n< \varepsilon$ for any $n \geq N$.
If you solve the inequality in $n$, you get $n > \log_{|r|}(\varepsilon)$, thus it is sufficient to take $N = \left\lceil\log_{|r|}(\varepsilon)\right\rceil$. 
(The direction of the inequality has been inverted because $x \mapsto \log_{|r|}(x)$ is a decreasing function for $|r|<1$.)
A: Write $r=\frac{1}{m}$ with $m>1$
Then $r^n=\frac{1}{m^n}$ But $m^n$ goes to infinity as $n$ goes to $\infty$ and so   $$r^n\rightarrow\frac{1}{\infty}=0$$
A: Hint: Let $r=\frac1m , m>1 \to$ you need to show $$lim_{n\to +\infty} r^n = 0$$
Since $0<r<1$
$$lim_{n\to +\infty} r^n = lim_{n\to +\infty} \left(\frac1m\right)^n=lim_{n\to +\infty}\frac{1}{m^n} = \frac{1}{+\infty} = 0$$
