$\infty\cdot 0$ Indetermination without L'Hopital? Evaluate $\lim_{n\to\infty}n\cdot r^n$ , being $0<r<1$.
I dont know if I took the proper steps, but I get to this point:
$$\lim_{n\to\infty}n\cdot r^n=\lim_{n\to\infty}n \lim_{n\to\infty}r^n = \infty \cdot  0 $$ 
I dont know how to solve this indetermination without L'hopital's rule.
 A: With $x_n=nr^n$, we have $\frac{x_{n+1}}{x_n}=\left(1+\frac{1}n\right)\cdot r$. Pickr $q$ with $1<q<\frac1r$. Then for $n\gg0$, $1+\frac1n<q$, hence $\frac{x_{n+1}}{x_n}<qr<1$. From this we conclude $x_n\to0$.
A: If
$$
n\ge n_r=\frac{\sqrt{r}}{1-\sqrt{r}}
$$
then
$$
1+\frac1n\le\frac1{\sqrt{r}}
$$
Therefore,
$$
\begin{align}
\frac{(n+1)r^{n+1}}{nr^n}
&=\left(1+\frac1n\right)r\\
&\le\sqrt{r}
\end{align}
$$
Which leads us to
$$
nr^n\le\overbrace{n_rr^{n_r}\vphantom{r^{\frac{n-n_r}2}}}^\text{constant}\overbrace{r^{\frac{n-n_r}2}}^{\to0}
$$
A: Hint: Substitute $$r=1/r'$$ then you will have $$\lim_{n\to \infty}\frac{n}{r'^n}$$
A: You want $\lim_{n\to\infty}n\exp -cn$ with $c:=-\ln r>0$. Since $\int_0^\infty n\exp -cn\operatorname{d}n=c^{-2}$ is finite, the limit is $0$.
A: $0<r<1.$
Set $r=\dfrac{1}{1+x}$, where $x >0$.
$(1+x)^n =$
$ 1+ nx + (1/2)n(n-1)x^2+......$
$\dfrac{n}{(1+x)^n} =$
$\dfrac{n}{1+nx +(1/2)n(n-1)x^2+...} < $
$\dfrac{2n}{n(n-1)x^2}<(2/x^2) \dfrac{n}{(n-1)^2} <$
$(2/x^2)\dfrac{n}{(n-n/2)^2}= (8/x^2)\dfrac {n}{n^2}=$
$(8/x^2)\dfrac{1}{n}.$
The limit is?
A: Although there are already many answers and an accepted one, I would like to add another one which also seems very elementary (most probably Sonnhard's hint points in this direction):
$$0 < r <1 \Rightarrow r = \frac{1}{1+q} \mbox{ with } q >0$$
For $n \geq 2$ we use binomial expansion:
$$(1+q)^n = 1 + nq + \color{blue}{\frac{n(n-1)}{2}q^2} + \cdots + q^n  \color{blue}{> \frac{n(n-1)}{2}q^2}$$
It follows immediately
$$nr^n = \frac{n}{(1+q)^n} < \frac{n}{\color{blue}{\frac{n(n-1)}{2}q^2}} = \frac{2}{(n-1)q^2} \stackrel{n \to \infty}{\longrightarrow} 0$$
A: HINT
We have that 
$$\ln (nr^n)=\ln n+n\ln r=n\cdot \left(\frac{\ln n}n+\ln r\right)\to -\infty$$
A: The series $\sum^\infty_1 x^{k+1}$ converges uniformly for $x\leq R<1$. Termwise differentiation applies for power series so $\sum^\infty_1 (k+1)x^k$ converges. In particular, $$\lim_{n\to\infty}nr^n\leq\lim_{n\to\infty}(n+1)r^n=0$$
