Define
$$V_n(\theta) = \sum_{k=-n-1}^{n+1} e^{in\theta} + \sum_{k=n+2}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta}+e^{-ik\theta})$$
I want to show that $$V_n(\theta) = 2K_{2n+1}(\theta) - K_n(\theta)$$ where $K_n(\theta)$ is the Fejer Kernel given by
$$K_n(\theta) = \sum_{k=-n}^{n}\left(1 - \frac{|k|}{n+1}\right)e^{ik\theta}$$
Expanding the definition of $K_n$, $$2K_{2n+1}(\theta) - K_n(\theta) = 2\sum_{k=-2n-1}^{2n+1}(1 - \frac{|k|}{2n+2})e^{ik\theta} - \sum_{k=-n}^n(1-\frac{|k|}{n+1})e^{ik\theta}$$
Then simplify the coefficients by finding a common denominator,
$$ \sum_{k=-2n-1}^{2n+1}\frac{2n+2-|k|}{n+1}e^{ik\theta} - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta} $$
I want to get some $(e^{ik\theta}+e^{-ik\theta})$ terms from the first summation,
$$ \frac{2n+2-0}{n+1}e^0 + \sum_{k=1}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta} + e^{-ik\theta}) - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$
Then I can break apart the first summation so that the indices match with $V_n$,
$$ 2 + \sum_{k=1}^{n+1}\frac{2n+2-k}{n+1}e^{ik\theta} + \sum_{k=n+2}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta} + e^{-ik\theta}) - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$
Great, the middle summation matches a term from $V_n$. Now my task seems easier, I just need to prove
$$\sum_{k=-n-1}^{n+1}e^{ik\theta} = 2 + \sum_{k=1}^{n+1}\frac{2n+2-k}{n+1}e^{ik\theta} - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$
and the "proof" is complete.
Unfortunately, this is where I get stuck. Futhermore, I'm not sure how to get rid of the $2$ term and this leads me to believe I've made an error somewhere. Can someone point out my error and put me or the right track?
