# Geometric series with arbitrary signs

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.

• 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 $\sum_{i=0}^{n-1} 2^i = 2^n-1$)
– Stef
Commented Apr 1, 2023 at 8:26

$$|S_n|$$ 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 $$\infty$$.

• 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. Commented Apr 1, 2023 at 12:22
• I know we're not supposed to pollute the comments with zero-content remarks, but ... Nice! Commented Apr 1, 2023 at 21:00
• 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. Commented Apr 6, 2023 at 9:40

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

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 $$|x| \lt 1$$.

• 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. Commented Apr 10, 2023 at 13:50
• 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. Commented Apr 11, 2023 at 1:42