How to prove that $x_n = nq^n$ for $|q| < 1$ is bounded?

Question

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.

Answer 1

Let fix $\epsilon>0$ then

$$|nq^n|\le \epsilon \iff\log n + n\log |q|\le\log\epsilon\iff\log n - n\log \frac 1{|q|}\le-\log \frac 1{\epsilon}$$

$$ n\log \frac 1{|q|}-\log n\ge \log \frac 1{\epsilon}\iff n\log \frac 1{|q|}-\log n+\log e\ge \log \frac 1{\epsilon}+ \log e$$

$$n\log \frac e{|q|}-\log n\ge \log n\log \frac e{|q|}-\log n \ge \log \frac e{\epsilon}$$

$$\log n\ge\frac{\log \frac e{\epsilon}}{\log \frac e{|q|}-1}$$

Answer 2

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.

Answer 3

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 $$0\leq\lim_{n\to \infty}\frac {n^p}{(1+r)^n}\leq \lim_{n\to \infty}\frac {n^p}{\binom {n}{k+1}}=0.$$

Answer 4

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).