Convergence of series $nq^n$. I've been faced with the following problem and would like some feedback.

Test the sequence $a_n = nq^n \space\space\space(-1 < q < 0)$ for convergence and determine the limit.

Now, my approach is using the Cauchy root test to prove that the corresponding series is convergent, which then implies $\lim_{n\to\infty}a_n = 0$.
$\forall n > N \space:\space\root_n\of{|a_n|} \le k < 1$
We'd end up with something like
$\root_n\of{|nq^n|} \le k < 1$, which I then separate into $\root_n\of n \space \cdot \space \root_n\of{|q^n|}$. The second part of that is effectively $|q|$ and the first one converges towards $1$, thus by my reasoning since $|q| < 1$, we can consider the series convergent, by the root criterion, which further implies that the sequence converges towards $0$.
(I am able to separately prove both of the additional claims that I used during this.)
Now my most obvious question would be if this is correct at all? The sequence is obviously convergent, but I'm not sure if I should even take this approach or possibly another one.
The second side question is that I'm not quite sure of the meaning of the constant $k$, and how the formulation above differs from 
$\forall n > N_0 \space:\space\root_n\of{|a_n|} < 1$?
Alternatively, I've also considered the ratio test
$$\forall n > N_0 \space:\space \left|\frac{a_{n+1}}{a_n}\right| \le k < 1$$
which leads me to
$\left|\frac{(n+1)q^{n+1}}{nq^n}\right|$ which simplifies to $\left|q + \frac{q}{n}\right|$.
By the triangle inequality we know that
$\left|q + \frac{q}{n}\right| \le |q| + \left|\frac{q}{n}\right|$
Now, since $|q| < 1$ and $\left|\frac{q}{n}\right|$ converges to $0$, it is indeed less than one, and thus convergent. Here too, however, I don't understand the role of $k$ and whether it invalidates my whole logic.
Thank you for your help!
 A: Compare the two series $\sum k^n$ with $|k|<1$ and $\sum \frac{n}{n+1}$.
In both cases the general term is positive and $<1$, yet the first serie converges, while the second one not.
Note that for $\sum \frac 1n$ the ratio test is precisely $\frac{n}{n+1}<1$ and the serie does not converge.
For the considered convergence tests, we want ultimately to make a comparison with a geometric serie, thus the request of a $k$ with $k<1$.
For instance with the ratio test : $|a_n|\le k|a_{n-1}|\le k^2|a_{n-2}|\le ... \le k^n|a_0|$
The partial sum is then
$|S_N|=|\sum\limits_{n=0}^{N} a_n|\le\sum\limits_{n=0}^{N} |a_n|\le(\sum\limits_{n=0}^{N} k^n)\;|a_0|$ which is a convergent serie.
The same arises for the root test since it is just a rewrite of $|a_n|\le k^n$ (in fact here the $a_0$ disappeared but this is simply because $\sqrt[n]{|a_0|}\to 1$ if $a_0\neq 0$).
With the $|q|+|\frac qn|$ thing, we can choose an epsilon such that 
$\begin{cases}
|q|\le 1-2\varepsilon & \text{from |q|<1} & \text{e.g. take }\varepsilon\le\frac{1-|q|}{2}\\
|q/n|\le\varepsilon & \text{from decreasing to 0}
\end{cases}$
Thus $|q+\frac qn|\le k$ with $0\le (k=1-2\varepsilon+\varepsilon= 1-\varepsilon)<1$ for $0<\varepsilon\ll 1$.
Note that $|\frac qn|\le\varepsilon$ happens only for a sufficiently large $n$ (in our specific case since |q|<1, it happens for any $n$, but let's generalize), however we are only interested in the tail of the serie since the partial sum is always finite anyway, so we can arbitrarily choose a large enough $n$.
