$\infty\cdot 0$ Indetermination without L'Hopital?

Question

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.

Answer 1

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 $$ nr^n\le\overbrace{n_rr^{n_r}\vphantom{r^{\frac{n-n_r}2}}}^\text{constant}\overbrace{r^{\frac{n-n_r}2}}^{\to0} $$

Answer 2

Hint: Substitute $$r=1/r'$$ then you will have $$\lim_{n\to \infty}\frac{n}{r'^n}$$

Answer 3

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 $x_n\to0$.

Answer 4

HINT

We have that

$$\ln (nr^n)=\ln n+n\ln r=n\cdot \left(\frac{\ln n}n+\ln r\right)\to -\infty$$

Answer 5

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 $0$.

Answer 6

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, $$\lim_{n\to\infty}nr^n\leq\lim_{n\to\infty}(n+1)r^n=0$$

Answer 7

$0<r<1.$

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?

Answer 8

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 $$nr^n = \frac{n}{(1+q)^n} < \frac{n}{\color{blue}{\frac{n(n-1)}{2}q^2}} = \frac{2}{(n-1)q^2} \stackrel{n \to \infty}{\longrightarrow} 0$$