16
$\begingroup$

Suppose $S_n = \sum_{i=0}^n c_i \alpha^i$, where $c_n \in \{ 1,-1\} $ for all $n \geq 0$, and $\alpha > 1$. I want to show that $|S_n| \to \infty$.

For $\alpha > 2$, it easily follows from the triangle inequality, $$|\sum_{i=0}^n c_i \alpha^i| \geq |c_n \alpha^n| - |\sum_{i=0}^{n-1} c_i \alpha^i| \geq |c_n \alpha^n| - \sum_{i=0}^{n-1} |c_i \alpha^i|.$$ For $\alpha = 2$, I managed to show this using uniqueness of binary representation of an integer. I am stuck with the case $1 < \alpha < 2$. I tried but could not find any counterexample. Any help will be appreciated.

$\endgroup$
1
  • $\begingroup$ In your last expression $|c_n \alpha^n| - \sum_{i=0}^{n-1} |c_i \alpha^i|$, all the $c_i$ and absolute values cancel out: $|c_n \alpha^n| - \sum_{i=0}^{n-1} |c_i \alpha^i| = \alpha^n - \sum_{i=0}^{n-1} \alpha^i = \alpha^n - \frac{\alpha^n - 1}{\alpha - 1} = \alpha^n \frac{\alpha - 2}{\alpha - 1}- \frac{1}{\alpha - 1}$. Which indeed goes to $+\infty$ if $\alpha > 2$, without mentioning "uniqueness of binary representation of an integer" (although the uniqueness of binary representation is more or less equivalent to the fact that $\sum_{i=0}^{n-1} 2^i = 2^n-1$) $\endgroup$
    – Stef
    Commented Apr 1, 2023 at 8:26

2 Answers 2

41
$\begingroup$

$|S_n|$ doesn't always limit to $\infty$. Consider the sum $$1+ \phi - \phi^2+ \phi^3 +\phi^4 -\phi^5 + \dots$$ Where $\phi$ is the golden ratio. This keeps returning to $0$, so $|S_n|$ can't limit to $\infty$.

$\endgroup$
3
  • 16
    $\begingroup$ Amazing. You used the root $\phi$ of $1+x-x^2$ and repeated the the sign pattern $++-$. But you can find a larger $\alpha$ with $1+x+x^2-x^3$ and pattern $+++-$. And in this way, by considering a root $\alpha$ of $1+x+x^2+...+x^{n-1}-x^n$ you can see that we can find an $\alpha$ that is as close to $\alpha=2$ as you wish. $\endgroup$ Commented Apr 1, 2023 at 12:22
  • 5
    $\begingroup$ I know we're not supposed to pollute the comments with zero-content remarks, but ... Nice! $\endgroup$
    – JonathanZ
    Commented Apr 1, 2023 at 21:00
  • 6
    $\begingroup$ Nice but in this case the sum simply has no limit at all. Instead it has three accumulation points, one is zero and the other two are at infinity. $\endgroup$ Commented Apr 6, 2023 at 9:40
0
$\begingroup$

This is not a proof but we give arguments for the following statement to hold

The sum

$$s = \sum_{i=0}^{\infty} c_{i} x^i$$

where $c_k \in [-1,+1]$ for $k=0,1,2,...$ converges only if $|x| < 1$.

Let us study the simple case of a periodic sign function, i.e. $c_{i+k}=c_i$ with $k=1,2,...$

Then for $k=2$ we have for the formally infinite sum (we write the more common $x$ instead of $\alpha$)

$$s_2 = \sum_{i=0}^{\infty} c_{i} x^i =\\c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... = \\c_0+ c_1 x + c_0 x^2 + c_1 x^3 ...=\\ (c_0 + c_1 x)(1+x^2 + x^4 + ...) \to \frac{(c_0 + c_1 x)}{1-x^2}$$

and for general $k$

$$s_k = \sum_{i=0}^{\infty} c_{i} x^i =(\sum_{i=0}^{k} c_i x^i)\left(1+x^k + x^{2k} + ...\right)=\frac{\sum_{i=1}^{k} c_i x^i}{1-x^k}$$

Hence the behaviour of the sum boils down to a finite sum times a geometric series. The convergence question is then easily answered and reads $ |x| \lt 1$.

For a non-periodic sign function this argument breaks down, of course. Hence it would be interesting to devise a non-periodic sign function and investigate that case.

Let us take the a non periodic function following the non-perodicity of a irrational number $z$.

Let $a_0, a_1, a_2, ...$ the binary representation of the number $z$.

Since we have $a_i \in [0,1]$, $c_i=2 a_i-1$ is a non-periodic sign function.

We now plot the sum $s$ as a function of $x$ for some famous irrational numbers

enter image description here

enter image description here

enter image description here

The emerging divergence at $|x| \to 1$ is manifest in all three cases.

This hints strongly towards a divergence of the sum for any distrubution of the signs when $|x| \to 1$.

In other words, the sum for abitrary $c_k$ converges only if $|x| \lt 1$.

$\endgroup$
2
  • $\begingroup$ There are a couple of typos in your formula for general $k$. Besides, the OP explicitly states $|a|>1$ (or $|x| > 1$ in your notation), so I fail to see how your post answers the original question. $\endgroup$
    – n_flanders
    Commented Apr 10, 2023 at 13:50
  • 1
    $\begingroup$ Thank you for pointing out the typos for general $k$. I have corrected them. And please notice that I have considered a general real variable $x$ instead of an α>1 . And then look to what size $|x|$ can grow for the sum to be convergent. The case of the OP is included in the non convergent region. $\endgroup$ Commented Apr 11, 2023 at 1:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .