The following proof of $c_n>0$ is based on Gjergji Zaimi's response to this [related question][1]. In particular the positive answer follows for that question, too. Moreover, the proof below should extend to a proof of $c_n>c_{n+2}$, but I was lazy to check this.

Let $n>0$ be even. By Chapter 6 of de Bruijn's "Asymptotic Methods in Analysis" (in particular by (6.4.6), (6.6.2), and the conclusion $P=0$ of Section 6.5), we have the following explicit formula:

$$ c_n = 2\pi^{-1/2}\sqrt{n!}\ \sum_{m=0}^\infty \ \int_{4m+1}^{4m+2} G_n(x)\ |\sin\pi x|^{-1/2}dx, $$

where

$$ G_n(x) := \sqrt{\frac{\Gamma(x)}{\Gamma(1+x+2n)}}- \sqrt{\frac{\Gamma(2+x)}{\Gamma(3+x+2n)}}.$$

It remains to show that $G_n(x)>0$ for $x\geq 1$. This reduces to

$$\Gamma(x)\Gamma(3+x+2n)>\Gamma(1+x+2n)\Gamma(2+x),$$

i.e. to 

$$ (1+x+2n)(2+x+2n)>x(1+x). $$ 

The last inequality is obvious, hence we are done.

  [1]: http://mathoverflow.net/questions/84958/is-sum-limits-n0-infty-xn-sqrtn-positive