Claim If $L_n=\displaystyle\frac{1}{\pi}\int_0^{\pi}|D_n(t)|dt$, we have $L_n=\dfrac{4}{\pi^2}\log n+O(1)$.
Proof We begin by dividing the interval $[0,\pi]$ into the subintervals where $\sin\left(n+\dfrac 1 2\right)$ keeps its sign. Since $\sin t/2$ is always non-negative in said domain, we may write $$\frac{1}{\pi }\int\limits_0^\pi {\left| {{D_n}\left( t \right)} \right|} = \frac{1}{\pi }\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} + \frac{1}{\pi }\sum\limits_{k = 1}^{n - 1} {\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} } + \frac{1}{\pi }\int\limits_{\frac{{2n\pi }}{{2n + 1}}}^\pi {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} $$ The last integral goes to $0$ as $n$ goes to infinity. We may focus our attention to the middle sum. The function $g(t)=\dfrac{1}{{\sin t/2}} - \dfrac{1}{t/2}$ is continuous on $[0,\pi]$, thus the integrals $${{c_n} = \frac 1 \pi\int\limits_0^{\frac{{2n\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|g\left( t \right)dt} }$$ exist and are bounded above by a constant for any $n$. We can then consider $$d_n=\frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{t}} } dt$$ But $t^{-1}$ is positive and decreasing, thus we have the bounds $$\frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\frac{{2n + 1}}{{2\left( {k + 1} \right)\pi }}\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} } dt \leqslant {d_n} \leqslant \frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\frac{{2n + 1}}{{2k\pi }}\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} } dt$$ But it is immediate that $${\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} }dt=\frac{4}{2n+1}$$ whence $$\frac{4}{{{\pi ^2}}}\sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \leqslant {d_n} \leqslant \frac{4}{{{\pi ^2}}}\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} $$
It is known that $\displaystyle\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} = \log n + O\left( 1 \right)$ (in fact, $=\log +n+\gamma+o(1)$), thus we can claim that ${d_n} = \dfrac{4}{{{\pi ^2}}}\log n + O\left( 1 \right)$, that is, $d_n-\dfrac{4}{\pi^2}\log n$ is bounded by a constant as $n\to\infty$. Finally, we look into the integral $$\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{t}} dt$$ that we excluded from our sum. The integrand is positive, continuous and decreasing on the interval in question, and reaches a maximum at $t=0$ with value $n+\dfrac 1 2$, whence we may give the crude estimation: $$\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{t}} dt \leqslant \left( {n + \frac{1}{2}} \right)\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {dt} = \pi $$
We know the sine rather well, that allows us to transform the integral
$$\int_{-\pi}^\pi \lvert D_n(x)\rvert\,dx$$
into something whose behaviour we can easier recognise. By symmetry, we need only consider the interval $[0,\pi]$. I don't like to type too many fractions either, so make the substitution $y = \frac{x}{2}$. Then we have
$$C_n := \int_{-\pi}^\pi \lvert D_n(x)\rvert\,dx = \frac{2}{\pi}\int_0^{\pi/2} \frac{\lvert\sin ((2n+1)y)\rvert}{\sin y}\,dy.$$
We can better analyse the behaviour if we get rid of the sine in the denominator. Replacing $\sin y$ with $y$ doesn't change much near $0$, where the most interesting things happen, so let's write
$$\frac{\pi}{2} C_n = \underbrace{\int_0^{\pi/2} \frac{\lvert \sin ((2n+1)y)\rvert}{y}\,dy}_{A_n} + \underbrace{\int_0^{\pi/2} \lvert \sin ((2n+1)y)\rvert \biggl(\frac{1}{\sin y} - \frac{1}{y}\biggr)\,dy}_{B_n}.$$
Since $\frac{1}{\sin y} - \frac{1}{y}$ is analytic in $\{ y : \lvert y\rvert < \pi\}$ and strictly positive on the open interval $(0,\pi)$, $B_n$ is positive and bounded above by $$B := \int_0^{\pi/2} \frac{1}{\sin y} - \frac{1}{y}\,dy.$$
If we care, it is not hard to show that $B_n \to \frac{2}{\pi}B$.
Now we can use that we know the zeros of the sine, and split the integral $A_n$ into parts between the zeros of $\sin ((2n+1)y)$. Let $z_k = \frac{k\pi}{2n+1}$, then
$$A_n = \int_0^{z_1}\frac{\sin ((2n+1)y)}{y}\,dy + \sum_{k=1}^{n-1} \int_{z_{k}}^{z_{k+1}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy + \int_{z_n}^{\frac{\pi}{2}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy.$$
We can neglect the last integral, since the denominator is bounded above by $1$ over the whole interval of integration there, and the length of the interval is $\frac{\pi}{2(2n+1)}$, so the contribution of that to $A_n$ is $O(n^{-1})$. The first integral is a constant independent of $n$, substituting $u = (2n+1)y$, we find
$$\int_0^{\frac{\pi}{2n+1}} \frac{\sin ((2n+1)y)}{y}\,dy = \int_0^\pi \frac{\sin u}{u}\,du.$$
We can now estimate
$$\frac{2}{(2n+1)z_{k+1}} < \int_{z_k}^{z_{k+1}} \frac{\lvert \sin ((2n+1)y)\rvert}{y}\,dy < \frac{2}{(2n+1)z_k}$$
using the strict monotonicity of $\frac{1}{y}$ and
$$\int_{z_k}^{z_{k+1}} \lvert \sin ((2n+1)y)\rvert\,dy = \frac{2}{2n+1}.$$
Summation yields
$$\frac{2}{\pi}(H_n - 1) < \sum_{k=1}^{n-1} \int_{z_{k}}^{z_{k+1}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy < \frac{2}{\pi} H_{n-1}$$
where $H_m = \sum\limits_{k = 1}^m \frac{1}{k}$ is the $m$-th harmonic number. As $H_m = \log m + \gamma + O(m^{-1})$ with the Euler-Mascheroni constant $\gamma$, we altogether obtain $A_n = \dfrac{2}{\pi}\log n + O(1)$, and therefore
$$C_n = \frac{4}{\pi^2}\log n + O(1).$$
Best Answer
Available in many places, the idea is to sum the $D_N$ using a geometric series fomula \begin{align} D_N(x) &= e^{-iNx}\sum_{k=0}^{2N} e^{ikx} \\ & =e^{-iNx}\frac{e^{(2N+1)x}-1}{e^{ix}-1} \\ &= \frac{e^{(N+1)x}-e^{-iNx}}{e^{ix}-1}\\ & = \frac{e^{(N+1/2)x}-e^{-i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}}\\ & = \frac{\sin((N+1/2)x)}{\sin(x/2)}\end{align} Alternatively, there's a nice telescoping sum argument here. Then $ \sum_{n=0}^N \sin((n+1/2)x) = \Im \sum_{k=0}^Ne^{i(n+1/2)x}$, and $$ \sum_{k=0}^Ne^{i(n+1/2)x} = e^{ix/2}\sum_{k=0}^Ne^{in x} = \frac{e^{i(N+1)x}-1}{e^{ix/2}-e^{-ix/2}} =\frac{e^{i(N+1)x}-1}{2i \sin(x/2)} $$ and therefore
$$F_N(x) = \frac1{(N+1)\sin(x/2)^2}\times (-1)\times \Re (\frac{e^{i(N+1)x}-1}2) =\frac{1-\cos((N+1)x)}{2(N+1)\sin(x/2)^2}$$ this already proves $F_N(x)\ge 0$, to put it in the "standard form" just use highschool trig $ \cos(2\theta) = 1 - 2\sin(\theta)^2$ to get $$ F_N(x) = \frac1{N+1} \left(\frac{\sin((N+1)x/2)}{\sin(x/2)}\right)^2\ge0.$$
For the convergence to zero, at every $x$ that is not a zero of $\sin(x/2)$, we have
$$ F_N(x) = \underbrace{\frac{1}{\sin(x/2)^2}}_{\text{constant in $N$}}\times \underbrace{\sin((N+1)x/2)^2}_{\le 1} \times \frac1{N+1}\to 0$$