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