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: $|nq^n|<\varepsilon$

HINT:
For $0<q<1$, let $q=1/(1+x)$ for $x>0$.  Then, write $nq^n=\frac{n}{(1+x)^n}$ and apply the binomial theorem. This should facilitate a $\delta-\epsilon$ proof.
Can you proceed now?
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

Using the binomial theorem we can write for $x>0$ \begin{align}(1+x)^n&=\sum_{k=0}^n\binom{n}{k}x^k\\\\&\ge \binom{n}{2}x^2\\\\&=\frac{n(n-1)}{2}x^2\end{align}Then, we have \begin{align}nq^n\le \frac{n}{\frac{n(n-1)}{2}x^2}=\frac{2}{(n-1)x^2}< \epsilon\end{align}whenever $n>1+\frac{2}{x^2\epsilon}$

