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$\{a_n\}$ is a bounded sequence (so it has upper and lower bound), $|q| < 1$. Prove that $s_n = a_0 + a_1 q + a_2 q^2 + \cdots + a_n q^n$ is a fundamental sequence (Cauchy sequence).

What I'll be writing next is not a proof, it's just a mere assessment, so I can show my "thinking process".
Well, let $b = \max \{a_n\}$. Then $$a_0 + a_1 q + a_2 q^2 + \cdots + a_n q^n \leq b + bq + bq^2 + \cdots + bq^n = \frac{b(1-q^n)}{1-q}$$ $q^n \to 0, n \to \infty$, so $$\frac{b(1-q^n)}{1-q} \to \frac{b}{1-q}$$ So we've got some fixed number $\frac{b}{1-q}$ (since $b$ and $q$ are also fixed numbers). At least, according to my assumption, it doesn't "converge" to infinity.
With $\varepsilon$-notation the sequence is fundamental, if $\forall \varepsilon$ $\exists N \in\mathbb{N}$ $\forall n \geq N$ $\forall p \geq 1:$ $ |a_{n+p} - a_n| < \varepsilon$
Then $$|s_{n+p} - s_n| \leq a_0 + a_1 q + a_2 q^2 + \cdots + a_{n+1} q^{n+1} + a_{n+2} q^{n+2} + \cdots + a_{n+p} q^{n+p} - a_0 - a_1 q - a_2 q^2 - \cdots - a_n q^n$$ $$|s_{n+p} - s_n| \leq a_{n+1} q^{n+1} + a_{n+2} q^{n+2} + \cdots + a_{n+p} q^{n+p}$$ Obviously, I can make the same move with $b = \max \{a_n\}$, but then: $$|s_{n+p} - s_n|\leq \frac{b(1-q^{n+p})}{1-q} - \frac{b(1-q^n)}{1-q} = \frac{b}{1-q}(q^n - q^{n+p})$$ Further we need to "extract" $n$ from expression $$\frac{b}{1-q}(q^n - q^{n+p}) < \varepsilon$$ in order to find $N$ for " $\forall \varepsilon$ $\exists N \in\mathbb{N}$ $\forall n \geq N$ $\forall p \geq 1:$ $ |a_{n+p} - a_n| < \varepsilon$ "
And while I'm sure that $1-q > 0$, I have no reason to assume that $b > 0$, so inequality sign is unknown in case of division both sides by $\frac{b}{1-q}$ Hence, something is wrong.

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It's not too hard to remedy this. All you have to do is be more careful with taking absolute values - note $q$ isn't necessarily positive either. Below, I treat $b$ as $\max|a_n|$, so that it is positive.

Fix $n,m\in\Bbb N$. A difference of partial sums: $$\begin{align}|s_{n+m}-s_n|&=|a_0+\cdots+a_{n+m}q^{n+m}-a_0-\cdots-a_nq^n|\\&\le|a_{n+1}q^{n+1}+\cdots+a_{n+m}q^{n+m}|\\&\le b\cdot\sum_{j=1}^m|q|^{n+j}\\&=b\cdot\frac{1-|q|^m}{1-|q|}\cdot|q|^{n+1}\\&<\frac{b}{1-|q|}\cdot|q^n|\end{align}$$

Now this last term is bounded independently of $m$, and as $|q|<1$ for all $\epsilon>0$ there will exist $N\in\Bbb N$ that $\frac{b}{1-|q|}|q|^n<\epsilon$ for all $n\ge N$. You can explicitly take: $$N:=\left\lceil\frac{1}{\log|q|}\log\left(\frac{\epsilon}{b}(1-|q|)\right)\right\rceil$$Where it is important to note that $\log|q|$ is negative.

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