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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 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

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

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

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

i.e. to

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

The last inequality is obvious, hence we are done.

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

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

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

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

i.e. to

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

The last inequality is obvious, hence we are done.

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

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

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

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

i.e. to

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

The last inequality is obvious, hence we are done.

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

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

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

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

i.e. to

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

The last inequality is obvious, hence we are done.

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.

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

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

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

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

i.e. to

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

The last inequality is obvious, hence we are done.