By application of Euler Identities we get that the Fourier Series of $f$ which is given by: $$f(\theta)\thicksim \sum_{n \ne o}\frac{(-1) ^ {n + 1}}{in} e^{in\theta} $$ is equal to $$2\sum_{n=1}^{\infty}(-1)^{n+1}\frac{sinn\theta}{n} $$
The book said that: It is possible to prove by elementary means that the above series converges for every $\theta$, but it is not obvious that it converges to $f(\theta)$, Could anyone explain this for me please?
Thanks.
Let us denote by $c_n$ the complex coefficients such that \begin{equation} c_n = \left\lbrace \begin{aligned} & \frac{(-1)^{n+1}}{\text{i} n}\!\!\!\! &\text{if}\quad n\neq 0\\ &0 &\text{if}\quad n= 0 \end{aligned} \right . = -c_{-n} \, . \end{equation} Therefore, using Euler identities, the series $\sum_{n\in\mathbb{Z}} c_n\, e^{\text{i}n\theta}$ rewrites as $\sum_{n\in\mathbb{N}} b_n \sin(n\theta)$, with the real coefficients $b_n = 2\text{i}\, c_n$.
Firstly, we examine the convergence of this series. One rewrites $\sum_{n\in\mathbb{Z}} c_n\, e^{\text{i}n\theta}$ as the value of a power series $\sum_{n\in\mathbb{Z}} c_n\, z^n$ at $z = e^{\text{i} \theta}$. Here, \begin{equation} \sum_{n\in\mathbb{Z}} c_n\, z^n = -\text{i}\log(1+z) + \text{i}\log(1+z^{-1}) \, , \end{equation} where $\log$ denotes the complex logarithm. The power series $\log(1+z)$ converges for all complex number $z\neq -1$ such that $|z|\leqslant 1$, and diverges like the harmonic series at $z=-1$. Therefore, the $2\pi$-periodic function \begin{equation} f : \theta \mapsto \sum_{n\in\mathbb{Z}} c_n\, e^{\text{i}n\theta} \end{equation} is defined on $\Omega = \mathbb{R}\setminus\lbrace \left(2k+1\right)\pi, k \in \mathbb{Z}\rbrace$.
Now, we consider a $2\pi$-periodic function $g$ with Fourier coefficients \begin{equation} \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t)\, e^{-\text{i} n t} \, dt = c_n \, , \end{equation} where $c_n$ is defined above. One sufficient condition for the Fourier series of $g$ to converge pointwise to the function $f$, i.e. $f(\theta) = g(\theta)$ for all $\theta$ in $\Omega$, is that $g$ is continuous and has left and right derivatives over $]-\pi, \pi[$. It is the case for $2\pi$-periodic functions $g$, which restriction to $]-\pi,\pi[$ is the identity.