As opposed to the sum in this question I want to prove the following:
Given $n\in \mathbb N$, $q\in \{\mathbb R \setminus0\}$ and: $$ x_n = \sum_{k=1}^n kq^{n-k} $$ Show that $x_n$ is an unbounded sequence.
Proof:
Expand $x_n$:
$$ x_n = q^{n-1} + 2q^{n-2} + 3q^{n-3} + \cdots+ (n-1)q^1 + nq^0 $$
Closed form for this sum may not be easily obtained in this form so try to transform it. Multiply $x_n$ by $(1-q^{-1})$:
$$ \begin{align} (1-q^{-1})\cdot x_n = q^{n-1} + &\color{red}{2q^{n-2}} + \color{green}{3q^{n-3}} + \cdots + \color{blue}{(n-2)q^2} + \color{orange}{(n-1)q^1} + nq^0 - \\ - &\color{red}{q^{n-2}} - \color{green}{2q^{n-3}} -\dots- \color{blue}{(n-3)q^2} - \color{orange}{(n-2)q^{1}} -(n-1)q^0-nq^{-1} \end{align} $$
Which after some transformations results in: $$ \begin{align} (1-q^{-1})\cdot x_n &= q^{n-1} + q^{n-2} + \cdots + q^1 + 1 - nq^{-1} = \\ &= \sum_{k=1}^{n-1}q^k + 1 - nq^{-1} =\\ &=\frac{q^n -q}{q-1} + 1 -nq^{-1} \tag1 \end{align} $$
Now if we multiply $(1)$ by $(q-1)$:
$$ (1-q^{-1})(q-1)\cdot x_n = q^n - 1 -n +nq^{-1} \implies \\ \implies x_n = \frac{q(q^n - q) -nq+n}{(q-1)^2} $$
Now consider $q^n$ and $nq$. Here $nq$ is obviously unbounded since $q$ is some given number such that $q \ne 0$. For $q^n$ consider the case when $0 < q < 1$. Let:
$$ q={ 1 \over 1+r } \\ r>0 $$ Then:
$$ q^n = {1\over (1+r)^n} $$
By Bernoulli:
$$ (1+r)^n \ge 1+nr \iff {1\over (1+r)^n} \le{1\over 1+nr} \implies \\ \implies q^n \le{1\over 1+rn} $$
If $q > 1$ then $q^n$ is unbounded. Therefore $x_n$ is an unbounded sequence.
End of proof.
Is it valid or have i missed something? Not sure whether I also need to prove for negative $q$, do I?
I forgot to mention that i'm only allowed to use precalculus methods. This problem is given even before the definition of limits.
