Convergence of a complex Fourier series

Question

Let $$ \sum_{k=-\infty}^\infty \frac{2}{\pi (2k+1)i} e^{(2k+1)it} $$

(*) For $n=2k$ the terms are zero.

I'd be glad for a guidance. How do I approach this? Should I split it for Real/Imaginary?

Answer 1

I don't think splitting into real and imaginary parts helps here. Rearranging and grouping each $e^{(2k+1)it},\, k\geqslant 0$ with $e^{-(2k+1)it}$ shows that

$$\sum_{k=-\infty}^\infty \frac{2}{\pi(2k+1)i}e^{(2k+1)it} = \sum_{k=0}^\infty \frac{2}{\pi(2k+1)i}\bigl(e^{(2k+1)it}-e^{-(2k+1)it}\bigr) = \frac{4}{\pi}\sum_{k=0}^\infty \frac{\sin \bigl(2k+1)t\bigr)}{(2k+1)}$$

is real, however.

To show convergence, in the original form we need to distinguish $t\in\pi\mathbb{Z}$ and $t\in\mathbb{R}\setminus \pi\mathbb{Z}$. For $t = m\pi$, we obtain

$$\sum_{k=-\infty}^\infty \frac{2(-1)^m}{\pi(2k+1)i}$$

which does not converge in the usual sense [independent convergence of the series for $k\geqslant 0$ and the series for $k < 0$], but the symmetric partial sums converge to $0$.

For $t\in\mathbb{R}\setminus \pi\mathbb{Z}$, the Dirichlet test shows convergence. In the rewritten real form, the distinction between $t\in\pi\mathbb{Z}$ and $t\notin \pi\mathbb{Z}$ need not be made, Dirichlet's test works for all $t$ then.

To find the function whose Fourier series this is, you could split the series and replace the $e^{it}$ with $z$, giving

$$\sum_{k=0}^\infty \frac{2}{\pi(2k+1)i} z^{2k+1} - \sum_{m=0}^\infty \frac{2}{\pi(2m+1)i}\overline{z}^{(2m+1)}.$$

The power series (one in $z$ and one in $\overline{z}$) have a radius of convergence $1$, we can explicitly determine their values for $\lvert z\rvert < 1$, and it is reasonable to hope that taking the limit $\lvert z\rvert\to 1$ gives the real function whose Fourier series we are given.

There are several ways to prove that the "reasonable hope" is indeed not misleading (that the method works in principle), but one can also verify it by computing the Fourier coefficients of the thusly determined function.