Notation clearification on Infinite Seires in Stein's Fourier Analysis? In the text "Stein and Shakarchi" Fourier Analysis I had conceptual troubles interrupting the notion of the following Series in 1.)
$$1.)\, \, \, \frac{1}{2i}\sum_{n\neq0}\frac{e^{inx}}{n}$$
Essentially my two key questions about the Series in 1.) is what does lower limit of $n \neq 0$ mean, and does the series have an upper limit ?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{1 \over 2\ic}\sum_{n \not= 0}{\expo{\ic nx} \over n} & =
{1 \over 2\ic}\sum_{n = 1}^{\infty}
\pars{{\expo{\ic nx} \over n} + {\expo{-\ic nx} \over -n}} =
\sum_{n = 1}^{\infty}{\sin\pars{nx} \over n} =
x\sum_{n = 1}^{\infty}\mrm{sinc}\pars{nx}
\\[5mm] & =
-x + x\sum_{n = 0}^{\infty}\mrm{sinc}\pars{nx} =
-x + x\bracks{\mrm{sgn}\pars{x}\,{\pi \over 2} + {1 \over 2}} =
\bbx{\ds{\pi\verts{x} - x \over 2}}
\end{align}
A: There are many ways to "sum over all nonzero integers". When the convergence is not absolute (as in this case) I think $$\displaystyle\sum_{n\neq0}$$ is ambiguous if not previously defined.
In this case it most likely means that it is the limit of partial symmetric sums$$\displaystyle\sum_{n\neq0}=\displaystyle\lim_{k\to\infty}\sum_{\substack{-k\leq n\leq k \\ n\neq0}}$$
With this definition your series converges at $x=0$ with value $0$.
