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5 Fixed some typos.

The following proof of $c_n>0$ is based on Gjergji Zaimi's response to this related question. In particular the positive answer follows for that question, too. Moreover, the argument below should also show that $c_n>c_{n+2}$.

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

$$G_n(x) := \sqrt{\frac{\Gamma(x)}{\Gamma(1+x+2n)}}- sqrt{\frac{\Gamma(x)}{\Gamma(1+x+n)}}- \sqrt{\frac{\Gamma(2+x)}{\Gamma(3+x+2n)}}.$$sqrt{\frac{\Gamma(2+x)}{\Gamma(3+x+n)}}.$$It remains to verify that G_n(x)>0 for x\geq 1. This reduces to$$\Gamma(x)\Gamma(3+x+2n)>\Gamma(2+x)\Gamma(1+x+2n),$$\Gamma(x)\Gamma(3+x+n)>\Gamma(2+x)\Gamma(1+x+n),$$

i.e. to

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

The last inequality is obvious, hence we are done.

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The following proof of $c_n>0$ is based on Gjergji Zaimi's response to this related question. In particular the positive answer follows for that question, too. Moreover, the proof argument below should extend to a proof of also show that $c_n>c_{n+2}$, but I was lazy to check this.c_n>c_{n+2}$. Let$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 verify that$G_n(x)>0$for$x\geq 1$. This reduces to $$\Gamma(x)\Gamma(3+x+2n)>\Gamma(1+x+2n)\Gamma(2+x),$$$\Gamma(x)\Gamma(3+x+2n)>\Gamma(2+x)\Gamma(1+x+2n),$$i.e. to$$ (1+x+2n)(2+x+2n)>x(1+x). $$The last inequality is obvious, hence we are done. 3 added 110 characters in body The following proof of c_n>0 is based on Gjergji Zaimi's response to this related question. In particular the positive answer follows for that question, too. Moreover, the proof below should extend to a proof of c_n>c_{n+2}, but I was lazy to check this. Let n>0 be even. By Chapter 6 of de Bruijn's "Asymptotic Methods in Analysis" (in particular by (6.4.6), (6.6.2), and the conclusion P=0 of Section 6.5), we have the following explicit formula:$$ 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.

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