# 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
Apr 1 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. Apr 1 at 12:22
• I know we're not supposed to pollute the comments with zero-content remarks, but ... Nice! Apr 1 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. Apr 6 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. Apr 10 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. Apr 11 at 1:42