Question: In an exercise I'm asked to study the limit of the following sequence:

$$(nr^n)\_{n\in \Bbb N}$$

I'm aware of the following:

* If $|r|>1$ then the sequence diverges if $r$ is positive, or it has no limit if $r$ is negative.
* If $r=1$ then it also diverges
* If $r = -1$ then it has no limit

But I'm not being able to find out what the limit is when $|r| < 1$. I know that if that's the case then $\lim\_n r^n = 0$, but I don't know how to do this for this sequence.

Answer: 
Assume $0 < r < 1$ and note that
$$n=n(r + (1-r))^n=nr^n\left(1 + \frac{1-r}{r}\right)^n.$$

Using the binomial theorem, we have

$$nr^n = \frac{n}{\left(1 + \frac{1-r}{r}\right)^n} < \frac{n}{1 + n\frac{(1 - r)}{r} + \frac{n(n-1)}{2}\frac{(1-r)^2}{r^2} }.$$

Therefore, $nr^n \to 0$ as $n \to \infty$.
Take

$$n\_0>\frac r{1-r}$$

so that for all $n\ge n\_0$
$$\frac{n+1}nr\le s<1.$$

Then by telescoping on $n\_0+1,n\_0+2,\cdots n\_0+n$,

$$\frac{n\_0+n}{n\_0}r^n<s^n$$ and $$nr^n<n\_0s^n.$$