Prove that $\lim_{n\to∞}(nq^n)=0$, while $\lvert q\rvert <1$ Prove that:
$$\lim_{n\to∞}(nq^n)=0$$ ($\lvert q\rvert <1$)
My only thought on this one is to treat q as a sub-series: $q=$ $1 \over b$ $\Rightarrow$ $q^n=$ $1\over b^n$ ;
So I can rewrite this as 
$$\lim_{n\to∞}({n \over b^n})=0$$ 
Any clues on how to solve this one?
I have to show that there is an $N \in \Bbb N$, and $\varepsilon >0$ which for every $n>N$ the following is true: $|nq^n|<\varepsilon$

The answer below was given before the OP changed his initial question.
It is sufficient to consider the case $0<q<1$. Then, if you know that, as $n \to +\infty$,
$$
\frac{\ln n}{n} \to 0
$$ then you may just consider
$$
\ln (nq^n)=\ln n+n\times \ln q=n \left(\underbrace{\ln q}_{\color{red}{<0}}+\underbrace{\frac{\ln n}{n}}_{\color{red}{\to 0}} \right) \to -\infty
$$ implying that
$$
nq^n \to 0, \quad n \to \infty.
$$
