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: $|nq^n|<\varepsilon$
 A: 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 $n>1+\frac{2}{x^2\epsilon}$

A: 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
$$
nq^n \to 0, \quad n \to \infty.
$$
A: ok, let's use the brute force approach here. To simplify my notation, I am going to assume non-negative $q$, but you can extend it easily to negative $q$ by substituting $q$ with $|q|$ below:


*

*let's find $N$ that satisfies $Nq^n < \epsilon$


$q^N < \epsilon /N < \epsilon$
$N =\ln \epsilon / \ln q$


*let's find $M$ after which $nq^n$ decreasing, that is:


$(M+1)q^{(M+1)} < Mq^M$
$q < \frac{M}{M+1} = 1 - \frac{1}{M+1}$
$\frac{1}{M+1} < 1- q$
$M > \frac{1}{1-q} -1$ (obviously $q < 1$)
That is for any $ n>M$, $nq^n$ has is smaller than its predecessor.


*Lastly let's choose $N^* = max(N, M)$. Every $n > N^*$ satisfies the $\epsilon $ inequality, so are all further values of $n$, thus you have proven your case.

