Observe that
$$\sum_{n=1}^\infty\frac{e^{in\theta}}n=\sum_{n=1}^\infty\frac{\cos n\theta}n+i\sum_{n=1}^\infty\frac{\sin n\theta}n$$
Splitting the sum above is justified because both series above converge, for example using Dirichlet's test (read Showing $\sum\frac{\sin(nx)}{n}$ converges pointwise and also http://mathforum.org/library/drmath/view/72101.html , for instance) , and this already proves convergence for $\;\theta\neq2k\pi\;,\;\;k\in\Bbb Z$ , since for $\;\theta=2k\pi\;$ we get the harmonic series.
Now, since for $\;z\in\Bbb C\;,\;\;|z|<1\;$ we have:
$$\frac1{1-z}=\sum_{n=1}^\infty x^{n-1}\implies -\text{Log}\,(1-z)=\sum_{n=1}^\infty\frac{z^n}n\;+\;K\text{ (=constant)}\implies$$
(Log$\,\,z\,$ is the complex logarithm) substitute $\;z=e^{i\theta}\;$ (this is justified by Abel's Theorem) :
$$-\text{Log}\,(1-e^{i\theta})=\text{Log}\,\frac1{1-e^{i\theta}}=\sum_{n=1}^\infty\frac{e^{in\theta}}n\;+\;K$$
Finally (fill in details of all the above), observe that
$$1-e^{i\theta}=1-\cos\theta-i\sin\theta\implies |1-e^{i\theta}|=\sqrt{2(1-\cos\theta)}$$ and also
$$\text{Log}\,z:=\log|z|+i\arg z\;\;,\;\;\text{with}\;\;\log\;\;\text{the real usual logarithm}$$
and usually choosing the main branch's principal value for the logarithm, in which $\;\arg z\in(-\pi,\,\pi]\;$ .
The summand $\;\frac i2F(\theta)\;$ is the constant above
You can start by integrating the trigonometric form of the Dirichlet kernel.
$$\int_{-\pi}^{\pi}D_N(x)dx=2\pi=\int_{-\pi}^{\pi}\frac{\sin((N+1/2)x)}{\sin(x/2)}dx$$
Rewrite the integrand to make use of the second fact from the prompt.
$$\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x}+\frac{2}{x})dx=2\pi$$
Split up the integral and consider the limit as $N\to\infty$.
$$\lim_{N\to\infty}\left[\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{1}{\sin(x/2)}-\frac{2}{x})dx+\int_{-\pi}^{\pi}\sin((N+1/2)x)(\frac{2}{x})dx\right]=2\pi$$
The first integral vanishes in the limit per the Riemann-Lebesgue lemma. The second has an even integrand, so take the part from zero to $\pi$ and simplify.
$$\lim_{N\to\infty}\int_{0}^{\pi}\sin((N+1/2)x)(\frac{1}{x})dx=\frac{\pi}{2}$$
Finally, change variables to $y=(N+1/2)x$.
$$\lim_{N\to\infty}\int_{0}^{(N+1/2)\pi}\frac{\sin(y)}{y}dy=\frac{\pi}{2}$$
Best Answer
\begin{align*} &L_{n}=\dfrac{1}{\pi}\int_{0}^{\pi}\left|\dfrac{\sin{(n+\frac{1}{2})x}}{\sin{\frac{x}{2}}}\right|dx>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left|\dfrac{\sin{(n+\frac{1}{2})2t}}{\sin{t}}\right|dt>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\dfrac{|\sin{(n+\frac{1}{2})2t}|}{t}dt\\ &=\dfrac{2}{\pi}\int_{0}^{(2n+1)\pi/2}\dfrac{|\sin{u}|}{u}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\dfrac{|\sin{u}|}{u}du\\ &=\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{0}^{\pi}\dfrac{\sin{u}}{u+k\pi}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\dfrac{1}{(k+1)\pi}\int_{0}^{\pi}\sin{u}du\\ &=\dfrac{4}{\pi^2}\sum_{k=0}^{n-1}\dfrac{1}{k+1} >\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma \end{align*} where $\gamma$ is Euler's constant. other hand
$$\left|\dfrac{\sin(n+1/2)t}{\sin{t/2}}-\dfrac{\sin{nt}}{\tan{t/2}}\right|\le 1$$
so \begin{align*} &L_{n}\le 1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\left|\dfrac{\sin{(2nt)}}{\tan{t}}\right|dt\\ &<1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\dfrac{|\sin{2nt}|}{t}dt\\ &=1+\dfrac{2}{\pi}\sum_{k=1}^{n}\int_{0}^{\pi}\dfrac{\sin{t}}{t+(k-1)\pi}dt\\ &<1+\dfrac{2}{\pi}\int_{0}^{\pi}\dfrac{\sin{t}}{t}dt+\dfrac{4}{\pi^2}\sum_{k=2}^{n}\dfrac{1}{k-1}\\ &<C+\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma \end{align*} so your result is obvious.