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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.

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  • $\begingroup$ Is $q$ always positive? According to the title, it can be negative too. $\endgroup$
    – Saša
    Oct 30, 2018 at 17:20
  • $\begingroup$ @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. $\endgroup$
    – roman
    Oct 30, 2018 at 17:21
  • $\begingroup$ Well, in that case, you should. Your proof looks valid but only for the half of possible values of $q$. $\endgroup$
    – Saša
    Oct 30, 2018 at 17:24

1 Answer 1

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

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