Here's problem 16 from Stein & Shakarchi's Fourier Analysis Chapter 5:
The Dirichlet kernel for periodic functions of period $1$ is defined by $$\mathcal{D}_R(x) = \widehat{\chi_{[-R,R]}}(x)=\frac{\sin(2\pi Rx)}{\pi x}.$$
Also, the modified Dirichlet kernel for periodic functions of period $1$ is defined by
$$\mathcal{D}^*_N(x) = \sum\limits_{|n|\leq N-1} e^{2\pi inx}+\frac{1}{2}\left(e^{-2\pi iNx}+e^{2\pi i Nx}\right)$$
Show that the result in Exercise 15 $\left(\sum\limits_{n=-\infty}^\infty \frac{1}{(n+\alpha)} = \frac{\pi}{\tan\pi\alpha}\right)$gives
$$\sum\limits_{n=-\infty}^\infty \mathcal{D}_N(x+n) = \mathcal{D}^*_N(x),$$
where $N\geq 1$ is an integer, and the infinite series must be summed symmetrically. In other words, the periodization of $\mathcal{D}_N$ is the modified Dirichlet kernel $\mathcal{D}^*_N$.
I couldn't figure out what the sentence in bold really meant and if it was needed for the solution.
I attempted a solution anyways, and it doesn't seem to be necessary:
Noticing that the numerator of the Dirichlet kernel (on the real line) is
$$\sin(2\pi N(x+n))= \sin(2\pi Nx+2\pi Nn) = \sin(2\pi Nx)\cos(2\pi Nn) + \sin(2\pi Nn)\cos(2\pi Nx) =\sin(2\pi Nx)\cdot 1 + 0 = \sin(2\pi Nx)$$ because $n,N \in \mathbb{Z}$, one gets $$\sum\limits_{n=-\infty}^\infty \frac{\sin(2\pi Nx)}{\pi(x+n)} = \frac{\sin(2\pi Nx)}{\pi}\sum\limits_{n=-\infty}^\infty \frac{1}{(x+n)}=_{15} \frac{\sin(2\pi Nx)}{\pi}\cdot\frac{\pi}{\tan\pi x} = \frac{\sin(2\pi Nx)}{\tan\pi x},$$
Now if only one could show that $\frac{\sin(2\pi Nx)}{\tan\pi x} = \mathcal{D}^*_N(x)$.
Using the complex exponential identities for $\sin$ and $\cos$, and that $\tan(x)=\frac{\sin(x)}{\cos(x)}$:
$$\frac{\sin(2\pi Nx)}{\tan\pi x} = \frac{e^{2\pi iNx}-e^{-2\pi i Nx}}{\require{cancel}\cancel{2i}}\cdot \frac{1}{\frac{e^{\pi ix}-e^{-\pi i x}}{\cancel{2i}}\cdot \frac{1}{\frac{e^{\pi ix}+e^{-\pi i x}}{2}}} = \frac{e^{2\pi iNx}-e^{-2\pi i Nx}}{e^{\pi ix}-e^{-\pi i x}}\cdot \frac{e^{\pi ix}+e^{-\pi i x}}{2}$$
$$= \left\{\left[\frac{e^{\pi i x(2N+1)}-e^{-\pi i x(2N+1)}}{e^{\pi i x}-e^{-\pi i x}}\right]+\left[\frac{e^{\pi i x(2N-1)}-e^{-\pi i x(2N-1)}}{e^{\pi i x}-e^{-\pi i x}}\right]\right\}\cdot \frac{1}{2}$$
letting $\xi = e^{2\pi i x}$,
$$e^{\pi i x(2N+1)}-\frac{1}{e^{\pi i x(2N+1)}} = \frac{e^{2\pi i x(2N+1)}-1}{e^{\pi i x(2N+1)}} = \frac{\xi^{(2N+1)}-1}{e^{\pi i x(2N+1)}}, e^{\pi i x(2N-1)}-\frac{1}{e^{\pi i x(2N-1)}} = \frac{e^{2\pi i x(2N-1)}-1}{e^{\pi i x(2N-1)}} = \frac{\xi^{(2N-1)}-1}{e^{\pi i x(2N-1)}}, e^{\pi i x}-\frac{1}{e^{\pi i x}} = \frac{e^{2\pi i x}-1}{e^{\pi i x}} = \frac{\xi -1}{e^{\pi i x}} $$
thus
$$\frac{\xi^{(2N+1)}-1}{e^{\pi i x(2N+1)}}\cdot\frac{e^{\pi i x}}{\xi -1} = \frac{\xi^{(2N+1)}-1}{\xi -1}\cdot \frac{1}{\xi^N}, \frac{\xi^{(2N-1)}-1}{e^{\pi i x(2N-1)}}\cdot \frac{e^{\pi i x}}{\xi -1} = \frac{\xi^{(2N-1)}-1}{\xi -1}\cdot \frac{1}{\xi^{N-1}}$$
using the geometric series
$\frac{1}{2}\cdot\left\{\frac{\xi^{2N}+\xi^{2N-1}+\cdots+1}{\xi^n}+ \frac{\xi^{2N-2}+\xi^{2N-3}+\cdots+1}{\xi^{N-1}}\cdot\frac{\xi}{\xi}\right\} =\frac{1}{2}\cdot\left\{\frac{\xi^{2N}+2\xi^{2N-1}+2\xi^{2N-2}+\cdots+2\xi+1}{\xi^N}\right\} $$ \\ \\ $$= \frac{1}{2}\left\{\xi^N +\xi^{-N} +2(\xi^{N-1}+\xi^{-(N-1)})+2(\xi^{N-2}+\xi^{-(N-2)})+2(\xi^{N-3}+\xi^{-(N-3)})+\cdots+2(\xi-\xi^{-1})+2\right\} $$ \\ \\ $$= \frac{1}{2}\cdot[\xi^{N}+\xi^{-N}] + \sum\limits_{|n|\leq N-1} \xi^{n} = \mathcal{D}^*_N(\xi)$
where $\xi = e^{2\pi i x}$. That is my attempt at a solution, is it correct? Even without the sentence in bold?