Proving a geometric sequence diverges. If I had some arbitrary sequence where $r > 1$ (ratio). How can I prove that it (being some $\{r^{n} \}$) diverges. I assume I can prove it using characteristics of geometric series, but I am unsure how to do so.
 A: $r=1+c$ where $c>0$.So by Bernouli's inequality:
$$r^n=(1+c)^n \geq 1+nc \to +\infty$$
A: Let $r=1+c,c\gt0$.  Then by the binomial theorem $r^n=(1+c)^n=\sum_{k=0}^n {n\choose k}c^{n-k}\ge1+nc\to \infty$.
A: Are you allowed to assume that $\ln N > 0$ if $N > 1$ and that $\ln M$ exists for every $M> 0$ and that for $0 < r \ne 1$ that  $\log_r M = \frac {\ln M}{\ln r}$ so that $r^{\log r M} = M$ is well defined?
If so the basic definition for $\lim_{n\to \infty} r^n = \infty$, means for any $M$ there is an $N$ so that $n > N$ implies $r^n > M$.  If $M > 1$ use $N=\log_r M$.  If $M \le 1$ use $N=1$.
If you can't use logarithms (it feasible to use limits to show logs are well define) the you can note that if $d = r-1>0$ then for any natural $n$ then $r^n = (1+d)^n = 1 + nd + \sum C_{n,d}d^n \ge 1+nd$. 
So for any $M$ let $N=\frac {M-1}d$.  If $n > N $ then $r^n \ge 1+nd \ge 1 + Nd = M$.
A: Another solution: Since $\dfrac{1}{r}>0$, then $\frac{1}{n}<\frac{1}{r}$ for almost all $n\in\Bbb N$. Then $r^n>n^n$ for almost all naturals $n$. Obviously $n^n\to\infty$.
