Please show me how to calculate the folowing limit of a sequence

I think should be something very simple... $$\lim_{n\to\infty}nq^{n}$$ wher $|q|<1$ i tried to use binomial theorem, but no success... I know that $q^{n}$ is convergent to $0$, but can't deal with this $n$

Hint: It is enough to deal with positive $q$. Let $q=e^{-k}$. We are interested in $$\lim_{x\to\infty} \frac{x}{e^{kx}}.$$ Now we can use L'Hospital's Rule.

If you want to use less machinery, let $q=\frac{1}{1+a}$. Use the Binomial Theorem to conclude that if $n\ge 2$ then $$(1+a)^n \ge 1+na+\frac{n(n-1)}{2}a^2\gt \frac{n(n-1)}{2}a^2.$$

Remark: We could simultaneously have dealt with negative $q$, by using absolute values appropriately. That would make things look more complicated than they are. For completeness, note that the case $q=0$ is trivial. And for negative $q$, we have $|nq^n|=n|q|^n$, so since $n|q|^n$ has limit $0$, so does $nq^n$.

Something fancy: the series

$$\sum_{n=0}^\infty x^n$$

has convergence radius equal to $\,1\,$ , so we can derive elementwise the above series and get a series with a convergence radius at least $\,1\,$:

$$\sum_{n-1}^\infty nx^{n-1}$$

Thus, for any $\,|q|<1\,$ , the series

$$\sum_{n=1}^\infty nq^n$$

converges and then $\,nq^n\xrightarrow [n\to\infty]{} 0\,$ for $\,|q|<1\,$

Here's another suggestion. Look at the ratio between successive terms, which is $$r_n=\left(1+\frac 1 n\right)q$$ If you look at the eventual behaviour of this from some suitable $N$, so that for $n\geq N, \left(1+\frac 1 n\right)<\frac 1 q$, then $r_n<r_N<1$ and for $k>0$ we have $$(N+k)q^{N+k} \leq r_N^kNq^N (=A_Nr_N^k \text { with constant } A_N)$$ and you can use the result you already have.

[The note in the answer given by André Nicolas about assuming positive $q$ is necessary to make this go through]