How to prove that $x_n = nq^n$ for $|q| < 1$ is bounded? This problem came up when i was solving another problem on boundedness.
Consider the following problem:

Let $n \in \mathbb N$ and
$$
\begin{cases}
x_n = nq^n \\
|q| < 1
\end{cases}
$$
Prove $\{x_n\}$ is a bounded sequence.

The problem above comes before the one I'm solving right now. I was unable to prove that $x_n$ is bounded and skipped that problem, but now I need to use it.
Context of the problem:

I'm working on proving that the following sum is bounded:
$$
\begin{cases}
y_n = \sum_{k=1}^nkq^k\\
|q| < 1
\end{cases}
$$

I've arrived at a closed form for the sum by expanding the terms and multiplying it by $(1-q)^2$, this is pretty easy to handle but takes a lot of space so i'm not posting it here. Here is the closed form for $y_n$:
$$
y_n = \frac{q(nq^{n+1} - q^n(n+1) + 1)}{(1-q)^2}
$$
So obviously the author of the problem expects me to first prove boundedness of $x_n$ before switching to $y_n$ because $y_n$ utilizes the prove on boundedness for $x_n$(note the $nq^{n}$).
My thoughts on proving boundedness for $x_n$:
For $x_n$ i have really no idea where even to start from. I've tried using Bernoulli's inequality and some tricks with binomial expansions but still couldn't handle it.
So my question is:

How to prove $x_n = nq^n$ is bounded. And can it be generalized for $z_n = n^pq^n$? For both cases $|q| < 1$ and for the second case $p\in \mathbb R$

Please note that this questions are precalculus ones, i'm not allowed to use calculus when solving it.
 A: Obvious for $q=0.$ If $0<|q|<1$ then let $|q|=\frac {1}{1+r}$ where $r>0.$ By the Binomial Theorem, if $n\in \Bbb Z^+$ and $r>0$ then $(1+r)^n\geq 1+nr.$
For general $p\in \Bbb R,$ with $r>0,$ take $k\in \Bbb Z^+$ with $p\leq  k.$ Then when $ k+1\leq n\in \Bbb Z^+$  we have by the Binomial Theorem that $$(1+r)^n\geq r^{k+1}\binom {n}{k+1}.$$ Now $r^{k+1}\binom {n}{k+1}$ is a polynomial in $n$ of degree $k+1$, with $k+1>p,$ so we have $$0\leq\lim_{n\to \infty}\frac {n^p}{(1+r)^n}\leq \lim_{n\to \infty}\frac {n^p}{\binom {n}{k+1}}=0.$$ 
A: I assume 
$q \ne 0; \tag 0$
with
$x_n = n q^n, \tag 1$
we have
$\dfrac{\vert x_{n + 1} \vert}{\vert x_n \vert} = \dfrac{\vert (n + 1)q^{n + 1} \vert }{\vert nq^n \vert} = \left \vert \dfrac{n + 1}{n}  \right \vert \vert q \vert = \left \vert 1 + \dfrac{1}{n} \right \vert \vert q \vert = \left ( 1 + \dfrac{1}{n} \right ) \vert q \vert; \tag 2$
since
$\vert q \vert < 1, \tag 3$
we have
$1 < \vert q \vert^{-1}, \tag 4$
whence for sufficiently large $n$
$1 < 1 + \dfrac{1}{n} < \vert q \vert^{-1}; \tag 5$
which yields
$\left ( 1 + \dfrac{1}{n} \right ) \vert q \vert < 1; \tag 6$
for such $n$, (2) implies
$\vert x_{n + 1} \vert = \left ( 1 + \dfrac{1}{n} \right ) \vert q \vert \vert x_n \vert < \vert x_n \vert, \tag 7$
which shows the sequence $x_n$ is bounded.  
If we take this a little further we can show the series
$y_n = \displaystyle \sum_{k = 1}^n kq^n \tag 8$
is bounded as well, since it may be written
$y_n = q^n \displaystyle \sum_{k = 1}^n k = \dfrac{n(n + 1)}{2} q^n; \tag 9$
as above, we form
$\dfrac{y_{n + 1}}{y_n} = \dfrac{ \dfrac{(n + 1)(n + 2)}{2} q^{n + 1}}{ \dfrac{n(n + 1)}{2} q^n} = \dfrac{n + 2}{n} q = \left ( 1 + \dfrac{2}{n} \right ) q; \tag{10}$
and again taking $n$ sufficiently large we have
$1 < 1 + \dfrac{2}{n} < \vert q \vert^{-1}, \tag{10}$
whence
$\left ( 1 + \dfrac{2}{n} \right )\vert q \vert < 1; \tag{11}$
thus, with $n$ large enough,
$\dfrac{\vert y_{n + 1} \vert}{\vert y_n \vert} < 1; \tag{12}$
that is,
$\vert y_{n + 1} \vert < \vert y_n \vert; \tag{13}$
therefore, the sequence $y_n$ is also bounded.  
In fact, both sequences $x_n$ and $y_n$ converge to $0$ as $n \to \infty$, since (6) and (11) each imply the existence of some $0 < \alpha < 1$ with
$\vert x_{n + 1} \vert < \alpha \vert x_n \vert, \tag{14}$
$\vert y_{n + 1} \vert < \alpha \vert y_n \vert, \tag{15}$
once $n$ becomes large enough.  It is easy to see that (14)-(15) imply, for some fixed but big enough $n$ that
$\vert x_{n + k} \vert < \alpha^k \vert x_n \vert \to 0 \; \text{as} \; k \to \infty, \tag{16}$
$\vert y_{n + k} \vert < \alpha^k \vert y_n \vert \to 0 \; \text{as} \; k \to \infty. \tag{17}$
Finally, consider the case of
$z_n = n^p q^n, \; p \in \Bbb R; \tag{18}$
as above we observe that
$\dfrac{\vert z_{n + 1} \vert}{\vert z_n \vert} = \dfrac{\vert (n + 1)^p q^{n + 1} \vert}{\vert n^p q^n \vert} =\dfrac{(n + 1)^p}{n^p} \vert q \vert = \left (1 + \dfrac{1}{n} \right )^p \vert q \vert; \tag{19}$
for $p \ge 0$ we may again take $n$ large enough that
$\left ( 1 + \dfrac{1}{n} \right )^p \vert q \vert \ < \alpha < 1; \tag{20}$
thus, as in (16)-(17),
$\vert z_{n + k} \vert < \alpha^k \vert z_n \vert \to 0 \; \text{as} \; k \to \infty; \tag{21}$
when $p < 0$,
$\left ( 1 + \dfrac{1}{n} \right )^p < 1, \; \forall n \ge 1, \tag{22}$
and so (20) binds for any $n \in \Bbb N$, and thus does (21).
A: hint
assume $q\ne 0$.
$$\ln(|x_n|)=n(\ln(|q|)+\frac{\ln(n)}{n})$$
thus
$$\lim_{n\to+\infty}|x_n|=0$$
a convergent sequence is bounded.
A: Let fix $\epsilon>0$ then
$$|nq^n|\le \epsilon \iff\log n + n\log |q|\le\log\epsilon\iff\log n - n\log \frac 1{|q|}\le-\log \frac 1{\epsilon}$$
$$ n\log \frac 1{|q|}-\log n\ge \log \frac 1{\epsilon}\iff n\log \frac 1{|q|}-\log n+\log e\ge \log \frac 1{\epsilon}+ \log e$$
$$n\log \frac e{|q|}-\log n\ge \log n\log \frac e{|q|}-\log n \ge \log \frac e{\epsilon}$$
$$\log n\ge\frac{\log \frac e{\epsilon}}{\log \frac e{|q|}-1}$$
