what is $\lim_{k \to 0}\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}$ where $A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$

Question

let $k>0$ and define $$A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$$

and

$$\gamma(k)=\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}.$$ Where $\Gamma$ is the Gamma function.

What is

$$\lim_{k \to 0^{+}} \gamma(k)?$$

First try

So

$$\gamma(k)=\frac{A-2kA+2A^3}{(k-A^2)^{\frac{3}{2}}}=\frac{A^{3}}{A^{3}} \frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}=\frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}$$

But I stuck in $\lim_{k \to 0}k/A^2$?

Since $\Gamma(\frac{1}{2})=\sqrt{\pi}$ so $A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})}$ so

$$\gamma(k)=\frac{\frac{\Gamma^2(\frac{k}{2})}{2\pi}-2k\frac{\Gamma^2(\frac{k}{2})}{2\pi}+2}{(k\frac{\Gamma^2(\frac{k}{2})}{2\pi}-1)^{\frac{3}{2}}}= \frac{ \frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{2\pi} }{ \frac{ (k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2} }{(2\pi)^{3/2}} } $$

$$=\sqrt(2\pi)\frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{(k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2}}$$

By using $\Gamma(z)=\frac1z - \gamma + o(1)$ (mentioned by @Brian Tung)

$$\Gamma^2(\frac{k}{2})=(\frac{2}{k} - \gamma + o(1))^2 =\frac{4}{k^2}+\gamma^2+o^2(1)-2\gamma \frac{2}{k} +2\frac{2}{k}o(1)-2\gamma o(1) $$

Second try

By knowing that $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$ [(4.1 The complement formula)][2]

$$A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})} =-\sqrt{2\pi} \frac{\Gamma(1-\frac{k}{2})\sin(\pi \frac{k}{2})}{\pi} \sim -\sqrt{2\pi} \frac{\sin(\pi \frac{k}{2})}{\pi}=-\sqrt{\frac{2}{\pi}} \sin (\frac{k\pi}{2}) $$

so by Taylor expansion $$A\sim -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} $$ and so

$$\gamma(k)=\frac{(1-2k)A+2A^3}{(k-A^2)^{\frac{3}{2}}}= \frac{(1-2k)\left( -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} \right)+2 \left( -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} \right)^3}{\left(k- \left( -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} \right)^2\right)^{\frac{3}{2}}} $$

Answer 1

Composing Taylor series, we have $$A=-\sqrt{2}\, \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$$ $$A=-\sqrt{\frac{\pi }{2}} k+\sqrt{\frac{\pi }{2}} k^2 \log (2)+O\left(k^3\right)$$ $$A^2=\frac{\pi k^2}{2}-\pi k^3 \log (2)+O\left(k^4\right)$$ $$A^3=-\frac{\pi ^{3/2} k^3}{2 \sqrt{2}}+\frac{3 \pi ^{3/2} k^4 \log (2)}{2 \sqrt{2}}+O\left(k^5\right)$$

So, (strangely written for me), $$(k+1)A-3kA+2A^3=-\sqrt{\frac{\pi }{2}} k+\sqrt{\frac{\pi }{2}} k^2 (2+\log (2))+O\left(k^3\right)$$ $$\left(k-A^2\right)^{3/2}=k^{3/2}-\frac{3}{4} \pi k^{5/2}+\frac{1}{2} k^{7/2} \left(\frac{3 \pi ^2}{16}+3 \pi \log (2)\right)+O\left(k^{9/2}\right)$$ Now, long division $$\gamma(k)=-\sqrt{\frac{\pi}{2 k}}\left(1+\frac{1}{4} k (-8+3 \pi -\log (16))+O\left(k^2\right)\right)\sim -\sqrt{\frac{\pi}{2 k}}\to -\infty$$

Answer 2

By using the expansion of $\gamma(k)$ we have: $$\gamma(k)=-\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{k}}+\frac{1}{8} \sqrt{k} \left(-3 \sqrt{2} \pi ^{3/2}-2 \gamma \sqrt{2 \pi }+8 \sqrt{2 \pi }-2 \sqrt{2 \pi } \psi ^{(0)}\left(\frac{1}{2}\right)\right)+O\left(k^{3/2}\right)$$ therefore $\gamma(k)\to -\infty$ as $k\to 0^+$