# Eulers Doubly Infinite Geometric Series

Recently I came across and "Identity" discovered by Euler.

$$E = \cdots + X^3 + X^2 + X + 1 + \frac{1}{X} + \frac{1}{X^2} + \frac{1}{X^3} + \cdots = 0$$

You can formally obtain this expression by applying the formula for the ordinary geometric series.

$$S = 1 + X + X^2 + \cdots = \frac{1}{1-X}$$

$$\Rightarrow E = \frac{1}{1-X} + \frac{1}{1-\frac{1}{X}} - 1 = 0$$

Disclaimer

Before the community jumps all over me and starts talking about,

• The radius of convergence of the geometric series
• Validity (or rather lack thereof) of manipulating divergent series.
• Some snarky statement about using this to prove 1=0.

Calm down, I know that the above is not a proof in $\mathbb{R}$ with the usual notion of convergence. And I am not claiming that it is. I know all about infinite series and partial sums and epsilons and deltas and all that great fantastic stuff.

This is just a neat mathematical exploration and I'm hoping some of you come along with me on this.

Motivation

Now the above "doubly infinite geometric series" actually has some uses. One thing is that it provides and insight in how to determine the representation of $-1$ as a p-adic number. For instance in the context of so called $10-adic$ numbers we can easily see from the arithmetic that,

$$\dots \overline{999}. + 1 = 0$$

Now in the reals,

$$1 = \sum_{k=1}^\infty \frac{9}{10^k}$$

$$\dots \overline{999} = \sum_{k=0}^\infty 9 \cdot 10^k$$

Which means that we could write,

$$0 = \dots \overline{999}. + 1 = \left( \sum_{k=0}^\infty 9 \cdot 10^k \right)_{\mathbb{Q}_{10}} + \left(\sum_{k=1}^\infty \frac{9}{10^k} \right)_{\mathbb{R}}$$

So it seems as if somehow this identity allows us to to relate the expressions for rationals in $\mathbb{Q}_p$ to the expressions for rationals in $\mathbb{R}$.

Observation

Something which has always struck me about the usual proof of the finite/infinite geometric series is that the formula depends only on the algebraic properties of the number system. In that sense perhaps we can think of this identity not as a statement about the real numbers (for which it is obviously not true) but somehow a statement about the properties of fields.

The Question

I want to know other contexts in which this "identity" has proven useful. Possible answers include but are not limited to,

• Any useful application of the identity distinct from the one above.

• Rigorous notions of convergence for which the identity makes sense.

• Rigorous settings in which the formal manipulations makes sense (e.g. similar to the way Algebraists think of polynomials, not as functions but algebraic entities).

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check for Laurent's series: [ en.wikipedia.org/wiki/Laurent_series ] –  janmarqz Feb 9 at 2:11
@janmarqz, thanks for the feedback! I am familiar with Laurent series in the context of complex analysis but I don't think that this particular one converges (except possibly somewhere on the unit disc?). I tried looking into formal Laurent series but it seems as though the convention is to only allow finitely many terms with a negative power to ensure that multiplication is well defined. Which of course excludes the identity. –  Spencer Feb 9 at 2:49
I doubt that this identity has any reasonable interpretation, in any reasonable topological ring. Taking $X=1$, we arrive at $\infty=0$. –  Martin Brandenburg Feb 10 at 1:20
@MartinBrandenburg, maybe but you never know until you find it. I've already found this identity to be a useful heuristic once and I am glad I didn't dismiss it out of hand. –  Spencer Feb 10 at 5:48
You could look at this as an analytic continuation argument in disguise. –  Cameron Williams Feb 10 at 14:56

We can look at this as a function inside of a function space. The function $\sum_{n=0}^\infty z^n = \frac{1}{1-z} = \phi(z)$ lies inside of the Smirnov Class. That is, it is a ratio of $H^2$ functions on the unit circle, $\mathbb{T} = \{ z \in \mathbb{C} : |z|=1\}$, with the denominator an outer function. In this case it is just a ratio of polynomials. That means it converges almost everywhere on the unit disc. There is one notable exception, $z=1$.

In this context we can view your series as $\phi(z) + \phi(\bar z) - 1$ for almost every $z$ in the unit circle. Which at least looks like it does converge to zero.

As long as you stay to the unit circle, you can do this same sort of analysis for a lot of functions with large coefficients.

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Thanks for taking a look at this. In light of Paul's observation above that would make this a reproducing kernel right? –  Spencer Feb 10 at 20:11
This isn't quite a reproducing kernel for the space that Paul mentioned. Given an orthonormal basis $e_n(z)$ in a RKHS we can construct the kernel function by $$K(z,w) = k_w(z) = \sum_{n\in\mathbb{Z}} \overline{e_n(w)}e_n(z).$$ In this case we have $e_n(z) = z^n \cdot (1+n^2)^{-s/2}$. Which gives $$K(z,w) = \sum_{n \in \mathbb{Z}} \bar w^n z^n (1+n^2)^{-s}.$$ Then you must verify that $k_w(z)$ is in fact in this space. –  Joel Feb 10 at 21:52
in that last summation do you mean $n \in \mathbb{Z}$? –  Spencer Feb 10 at 21:54
Yes indeed. I edited it. –  Joel Feb 10 at 21:54
They really need a preview for the Latex compilation in these comments –  Spencer Feb 10 at 21:55
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Letting $z=e^{i\theta}$, the Fourier expansion ${1\over 2\pi} \sum_{n\in \mathbb Z} 1\cdot e^{in\theta}$ is the Fourier expansion of a "Dirac comb", meaning a sum of Dirac delta distributions at integer multiples of $2\pi$. This Fourier series converges perfectly well in a Sobolev-like space with Hilbert-space norm $$\Big|\sum_n c_n\,e^{in\theta}\Big|_s^2 \;=\; \sum_n |c_n|^2\cdot (1+n^2)^s$$ for $s<-1/2$. Indeed, for sufficiently nice functions, the fact that their Fourier series converge pointwise to them can be interpreted as verifying that this distribution behaves as claimed.

In particular, the fact that this distribution really is locally $0$ away from the Dirac-delta spikes is a sort of certification that it "is zero" there. (Distributions are localizable.)

That is, the Fourier series of (periodic) distributions certainly do not converge pointwise, just as Fourier transforms of tempered distributions often do not. But such things do certainly converge in a different topology, and are enormously useful.

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Nice answer. I should have thought of a weighted Sobolev Space. –  Joel Feb 10 at 15:23
@paulgarret, That makes a lot of sense. Thank you for taking the time to answer this. –  Spencer Feb 10 at 20:16