$\lim_n nr^n$ if $|r| < 1$

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 1

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 $n \to \infty$.

Answer 2

From the ratio test you know that $n r^n \to 0$ if:

$$\lim_{n \to \infty} \left|\frac {(n + 1)r^{n + 1}} {n r^n}\right| < 1$$

Alternatively you can use the root test, or consider the derivative of $(1 - x)^{-1} = \sum x^n$ as suggested by Ethan Bolker in the comments, though this is pretty non-trivial if you're to justify it properly.

Answer 3

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 $$nr^n<n_0s^n.$$