Why is the Fejér Kernel always non-negative? In one of the proofs in my notes the Fejér Kernel ($K_{n}$ below) is plucked out of an absolute value seemingly for free.  On the previous page it is remarked that this function is always non-negative.  I can't see why this would be automatic.  Why is this true?
We have $K_{n} = \sum\limits_{j=-n}^{n} D_{j}(t)$, where $D_{n} = \sum\limits_{j=-n}^{n}e^{ijt}$.
It is not automatic that $e^{ijt}$ is non-negative (or even real) for every value of $t\in [-\pi,\pi)$, $j\in\mathbb{Z}$.
This is what confuses me.
 A: First, notice that 
$$
D_n(t) = \sum_{k=-n}^n \exp(i k t) = \exp(-i t n) \sum_{k=0}^{2n} \exp(i k t) = \mathrm{e}^{-i t n} \frac{1-\exp(i t (2n+1))}{1-\exp(i t)} = \frac{\sin ( n t + \frac{t}{2})}{\sin(\frac{t}{2})}
$$
Then by definition $K_{n} = \frac{1}{n} \sum_{k=0}^{n-1} D_k(t)$ we have:
$$ \begin{eqnarray}
 K_n &=& \frac{1}{n \sin(t/2)} \sum_{k=0}^{n-1} \sin\left( k t+\frac{t}{2}\right) \\
   &=& \frac{1}{n \sin(t/2)^2} \frac{1}{2}\sum_{k=0}^{n-1} \left( \cos(k t) - \cos((k+1)t)\right)   \\
   &=& \frac{1}{2 n \sin(t/2)^2} \left( 1 - \cos( n t) \right) =
     \frac{1}{n} \left( \frac{\sin(n t/2)}{\sin(t/2)} \right)^2 \ge 0
  \end{eqnarray}
$$
A: Using the formula for the partial sum of a geometric series, the Dirichlet kernel is
$$
D_n(t)=\frac{e^{i(2n+1)t/2}-e^{-i(2n+1)t/2}}{e^{it/2}-e^{-it/2}}\tag{1}=\frac{\sin((2n+1)t/2)}{\sin(t/2)}
$$
We will sum the middle form, again using the formula for the partial sum of a geometric series, to get
$$\begin{align}
F_N(t)
&=\frac{1}{N}\sum_{n=0}^{N-1}D_n(t)\\
&=\frac{1}{N}\frac{e^{iNt}-2+e^{-iNt}}{(e^{it/2}-e^{-it/2})^2}\\
&=\frac{1}{2N}\frac{1-\cos(Nt)}{\sin^2(t/2)}\\
&=\frac{1}{N}\frac{\sin^2(Nt/2)}{\sin^2(t/2)}\tag{2}
\end{align}
$$
A: You can prove that
$$\frac{1}{n}\sum_{j=0}^{n-1} D_j(t)=\begin{cases} \frac{1-\cos{nt}}{(1-\cos{t})n} &\mbox{if } t\not=2k\pi \\
n & \mbox{if } t=2k\pi \end{cases}\ge0$$
by induction on $n$ if $t=2k\pi$ and by multiplying $D_j(t)$ by $e^{it}-1$ otherwise (a telescopic series will appear).
