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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 ?

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    It's the sum over all nonzero integers. – 2017-02-24
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    Ahh, thank you I Initially thought that at first just wanted to verify. – 2017-02-24

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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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 $0$.

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    Can you provide me with the intuition behind notation of Symmetric Summation ? I do not have an understanding of your solution. – 2017-02-27
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    @Zion Usually, if we have numbers $(a_k)$ $(k\in\Bbb{Z})$, we wriite $$\sum_{k=-\infty}^{\infty}a_k=L$$ if $$\forall\epsilon>0,\exists N,\forall m,n>N,\left|\sum_{k=-m}^n a_k-L\right|<\epsilon\tag{1}$$ For me a Fourier series is a "sum over all integers" in this sense. However some results in Fourier analysis concern only *symmetric* partial sums. In your case, it seems like the notation $$\sum_{k=-\infty}^{\infty}a_k=L$$ means $$\lim_{n\to\infty}\sum_{k=-n}^n a_k=L$$ that is to say $$\forall\epsilon>0,\exists N,\forall n>N,\left|\sum_{k=-n}^n a_k-L\right|<\epsilon\tag{2}$$ – 2017-02-27
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    @Zion The definition $(1)$ is more restrictive than $(2)$. It might happen that the limit exists according to $(2)$ but not according to $(1)$. However, if the limit exists according to $(1)$ then it exists according to $(2)$ and the limit values are the same. When convergence is absolute according to one of the two definitions, then the definitions coincide. That's basically a reflection of the fact that when convergence is absolute the order of summation plays no role. – 2017-02-27
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    @Zion A definition of $$\sum_{k=-\infty}^{\infty}a_k=L$$ equivalent to $(1)$ and in terms of usual "one-sided" series is $$\sum_{k=0}^{\infty}a_k\text{ and }\sum_{k=1}^{\infty}a_{-k}\text{ both exist and }\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}=L$$ that is to say $$\sum_{k=-\infty}^{\infty}a_k = \sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}$$ – 2017-02-27
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    @Zion By the way this is similar to the definition of Riemann integrals of the type $\displaystyle\int_{-\infty}^{\infty}f$ which are defined as $$\int_{-\infty}^{\infty}f = \lim_{m,n\to\infty}\int_{-m}^nf$$ or $$\int_{-\infty}^{\infty}f = \int_{-\infty}^{0}f + \int_{0}^{\infty}f $$ (of course the choice of break point $0$ is arbitrary, we could take any other real number). The *symmetric* analog $$\lim_{n\to\infty}\int_{-n}^nf$$ is called the [Cauchy principal value](https://en.wikipedia.org/wiki/Cauchy_principal_value). The latter might exist even if $\int_{-\infty}^{\infty}f$ doesn't. – 2017-02-27