# Proof verification of $\sum_{k=1}^n kq^{n-k}$ is unbounded for $n \in \mathbb N$ and $q\in\{\mathbb R\setminus0\}$

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.

• Is $q$ always positive? According to the title, it can be negative too.
– Saša
Oct 30, 2018 at 17:20
• @Oldboy, you are right, it may be negative. That's why i've put a remark doubting whether i should consider that case as well. Oct 30, 2018 at 17:21
• Well, in that case, you should. Your proof looks valid but only for the half of possible values of $q$.
– Saša
Oct 30, 2018 at 17:24

hint write $$x_n$$ as $$x_n=\sum_{k=1}^n(n-k)q^k$$