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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.

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  • $\begingroup$ In my case $r=\frac{1}{5}$ $\endgroup$ – user605734 MBS Nov 2 '18 at 15:03
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    $\begingroup$ \cdot produces $\cdot$ for multiplication, whereas . is just a decimal point. $\endgroup$ – Asaf Karagila Nov 2 '18 at 15:06
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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 $$ nr^n\le\overbrace{n_rr^{n_r}\vphantom{r^{\frac{n-n_r}2}}}^\text{constant}\overbrace{r^{\frac{n-n_r}2}}^{\to0} $$

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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 $x_n\to0$.

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

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  • $\begingroup$ How can turning a $\infty \cdot 0$ indetermination into a $\frac{\infty}{\infty}$ help? $\endgroup$ – user605734 MBS Nov 2 '18 at 15:32
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    $\begingroup$ $n$ increase linear but $$r'^n$$ like a potence function $\endgroup$ – Dr. Sonnhard Graubner Nov 2 '18 at 15:36
  • $\begingroup$ 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? $\endgroup$ – user605734 MBS Nov 2 '18 at 15:47
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    $\begingroup$ Yes $$5^n$$ grows faster than $n$, and yes the Limit is $0$ $\endgroup$ – Dr. Sonnhard Graubner Nov 2 '18 at 15:53
  • $\begingroup$ I feel as though showing that $5^n$ grows fast than $n$ is exactly the content of this limit. $\endgroup$ – Santana Afton Nov 2 '18 at 18:04
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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 $$nr^n = \frac{n}{(1+q)^n} < \frac{n}{\color{blue}{\frac{n(n-1)}{2}q^2}} = \frac{2}{(n-1)q^2} \stackrel{n \to \infty}{\longrightarrow} 0$$

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HINT

We have that

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

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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 $0$.

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

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$0<r<1.$

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?

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