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

Evaluate $$\lim_{n\to\infty}n\cdot r^n$$ , being $$0.

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.

• In my case $r=\frac{1}{5}$ – user605734 MBS Nov 2 '18 at 15:03
• \cdot produces $\cdot$ for multiplication, whereas . is just a decimal point. – Asaf Karagila Nov 2 '18 at 15:06

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

With $$x_n=nr^n$$, we have $$\frac{x_{n+1}}{x_n}=\left(1+\frac{1}n\right)\cdot r$$. Pickr $$q$$ with $$1. Then for $$n\gg0$$, $$1+\frac1n, hence $$\frac{x_{n+1}}{x_n}. From this we conclude $$x_n\to0$$.

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

• How can turning a $\infty \cdot 0$ indetermination into a $\frac{\infty}{\infty}$ help? – user605734 MBS Nov 2 '18 at 15:32
• $n$ increase linear but $$r'^n$$ like a potence function – Dr. Sonnhard Graubner Nov 2 '18 at 15:36
• Then with, for example, $r=\frac{n}{5^n}$ I just have to argue that $5^n$ grows faster than $n$, then the limit is 0? – user605734 MBS Nov 2 '18 at 15:47
• Yes $$5^n$$ grows faster than $n$, and yes the Limit is $0$ – Dr. Sonnhard Graubner Nov 2 '18 at 15:53
• I feel as though showing that $5^n$ grows fast than $n$ is exactly the content of this limit. – Santana Afton Nov 2 '18 at 18:04

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

HINT

We have that

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

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 $$0$$.

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

$$0

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?