Questions about delta function.

Question

Let $z \neq 1$ be a complex number. Then \begin{align} \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n. \end{align}

We have \begin{align} \frac{z^{-1}}{1-z^{-1}} = \sum_{n=1}^{\infty} z^{-n}. \end{align}

If we add the left hand sides of the two equations, then we have \begin{align} \frac{1}{1-z} + \frac{z^{-1}}{1-z^{-1}} = 0. \end{align} If we add the right hand sides of the two equations, then we obtain \begin{align} \sum_{n = - \infty}^{\infty} z^n. \end{align} Therefore when $z \neq 1$, we have $\sum_{z=-\infty}^{\infty} z^n= 0$. Do we have $\delta(z) = \sum_{n=-\infty}^{\infty} z^n$? How to understand the delta function and analytic continuation? Any help will be greatly appreciated!

Answer 1

You can get something like that only in some kind of distributional context. As distribution, you can write, as one version $$ \sum_{n\in\Bbb Z}z^n=\lim_{q\to 1-0}\sum_{n\in\Bbb Z}q^{|n|}z^n =\lim_{q\to 1-0}\frac{1-q^2}{1-q(z+z^{-1})+q^2} $$ which for $|z|=1$ and away from $z=1$ converges to zero while the integral over the unit circle always evaluates to $2\pi i$. Which is the condition for any sequence approximating (a multiple of) the delta distribution.

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In the end, regardless of how one sets up the test functions and pairing with them, this identity amounts to one classical fact:

For smooth enough periodic functions, the Fourier series converges pointwise to the function values.

The pairings with the $z^n$ produce (multiples of) the Fourier series coefficients, and the series summation gives the evaluation of the Fourier series at $z=1$ resp. $t=0$ for $z=e^{it}$.

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Standard reference for this topic: Carl Offner: A little harmonic analysis

Answer 2

The first of your series converges for $|z| < 1$, the second for $|z| > 1$.

In other words, there are no values of $z$ for which it is meaningful to add the two series.