p-norm of series / Mathematica When computing the $p$-norm of the series $x_n = \mathrm e^{\mathrm i n}$ for $n \in \mathbb{Z}$ (not $\mathbb{N}$), the sum should not converge:
$$\|x\|_p^p = \sum_{n \in \mathbb{Z}} (\mathrm e^{\mathrm i n})^p = \sum_{n \in \mathbb{Z}} (\mathrm e^{\mathrm i p})^n$$ and $|\mathrm e^{\mathrm i p}|=1$, so the series diverges. 
For $p = 1$ this is also the result of Mathematica and Wolfram Alpha.
However, using general $p$, the result of Mathematica and Wolfram Alpha is 0. This is strange. What am I missing here?
 A: Your series is indeed divergent but in Fourier series (and distribution theory) we may formally write : $$  \delta(x)=\frac1{2 \pi} \sum_{n=-\infty}^{\infty} e^{i nx}$$ with $\delta$ the Dirac Delta distribution 
From this point of view your infinite sum will be $2\pi\delta(x)$ that is $0$ for $x\ne 0$ and $\infty$ for $x=0$.
Another related form of the Dirac distribution is this integral :
$$  \delta(x)=\frac1{2 \pi}\int_{-\infty}^{\infty} e^{i t x} dt$$
The Fourier inversion theorem may help too (less distribution theory required!).

EDIT: Coming back to your second Alpha link I should add that (the first time) I got :
$$\sum_{n=-\infty}^{\infty} \exp^p(i n) = \frac{(e^i)^p-e^{i p}}{(-1+(e^i)^p) (-1+e^{i p})}$$
After that I got $0$ and indeed the numerator of this answer is $0$.
I don't know the rules used by Mathematica to get this (perhaps a limited sum process turning wrong or right or whatever at the limit!) but anyway these very powerful tools should not be considerer as references!
 They make assumptions (for example that your variables are complex numbers), apply rules that may be conflicting (real or complex analysis versus germs of distribution theory) and simplify at each stage with perhaps inapplicable methods (during a computation of limit for example). So that, sometimes, you'll get discussable answers... especially when considering divergent expressions! 
