Computing $\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}+1\right)}{k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$ in closed form What tools other than beta function you might like to use here?
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^2 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.27541$$
Supplementary question: calculating 
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^3 \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}\approx 1.02593$$
Is there a way to generalize it and get such a calculation? 
$$\sum _{k=1}^{\infty } \frac{\displaystyle \Gamma \left(\frac{k}{2}+1\right)}{\displaystyle k^n \Gamma \left(\frac{k}{2}+\frac{3}{2}\right)}$$

Writing $\dfrac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}$ as an integral and exchanging summation and integration, we get
\begin{align}
\sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}
&=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x{\rm Li}_2(x)}{\sqrt{1-x^2}}\ {\rm d}x\tag1\\
&=-\frac{2}{\sqrt{\pi}}\int^1_0\frac{\sqrt{1-x^2}\ln(1-x)}{x}\ {\rm d}x\tag2\\
&=-\frac{1}{\sqrt{\pi}}\int^\pi_0\frac{\cos^2{x}\ln(1-\sin{x})}{\sin{x}}\ {\rm d}x\tag3\\
&=-\frac{1}{\sqrt{\pi}}\left[\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\mathcal{A}_n+2\sum^\infty_{n=0}\frac{(-1)^{n-1}}{2n+1}\mathcal{B}_n\right]\tag4\\
\end{align}
where 
\begin{align}
\mathcal{A}_n
=\int^\pi_0\frac{\cos^2{x}}{\sin{x}}(\cos(2nx)-1)\ {\rm d}x
\ \ \ , \ \ \ \mathcal{B}_n
=\int^\pi_0\frac{\cos^2{x}}{\sin{x}}\sin((2n+1)x)\ {\rm d}x\\
\end{align}
Using simple trigonometric identities, it is not hard to see that, for $n\in\mathbb{N}$,
\begin{align}
\mathcal{A}_n-\mathcal{A}_{n-1}&=-\ \frac{1}{2n-3}-\frac{2}{2n-1}-\frac{1}{2n+1}\tag5\\
\mathcal{B}_n-\mathcal{B}_{n-1}&=0\tag6
\end{align}
Thus we may obtain the closed forms for both sequences.
\begin{align}
\mathcal{A}_n=2H_n-4H_{2n}+\frac{1}{2n-1}-\frac{1}{2n+1}+2\ \ \ , \ \ \ \mathcal{B}_n=\pi-\frac{\pi}{2}\delta_{n0}\tag7
\end{align}
Using the Taylor series for $\ln(1-x)$ and $\arctan{x}$, as well as the well-known identities
\begin{align}
\sum^\infty_{n=1}\frac{(-1)^{n-1}H_n}{n}&=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\tag8\\
\sum^\infty_{n=1}\frac{(-1)^{n-1}H_{2n}}{n}&=\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\tag9
\end{align}
gives us
\begin{align}
\sum^\infty_{k=1}\frac{\Gamma\left(\frac{k}{2}+1\right)}{k^2\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}
&=-\frac{1}{\sqrt{\pi}}\left[2\left(\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}\right)-4\left(\frac{5\pi^2}{48}-\frac{1}{4}\ln^2{2}\right)+2-2\pi\left(\frac{\pi}{4}\right)+2\left(\frac{\pi}{2}\right)\right]\\
&=\frac{3\pi^2-4\pi-8}{4\sqrt{\pi}}\tag{10}
\end{align}
as the closed form.

Explanation: 
$(1)$: Write $\displaystyle\frac{\Gamma\left(\frac{k}{2}+1\right)}{\Gamma\left(\frac{k}{2}+\frac{3}{2}\right)}=\frac{2}{\sqrt{\pi}}\int^1_0\frac{x^{k+1}}{\sqrt{1-x^2}}\ {\rm d}x$. 
$(2)$: Integrated by parts. 
$(3)$: Substitute $x\mapsto\sin{x}$ then $x\mapsto\pi-x$. 
$(4)$: Use the fact that $\ln(2-2\sin{x})=2\mathrm{Re}\ln(1+ie^{ix})$ then expand the $\mathrm{RHS}$. 
$(5)$: Write $\displaystyle\mathcal{A}_n-\mathcal{A}_{n-1}=-\int^\pi_0(1+\cos(2x))\sin((2n-1)x)\ {\rm d}x$. 
$(6)$: Write $\displaystyle\mathcal{B}_n-\mathcal{B}_{n-1}=\int^\pi_0(1+\cos(2x))\cos(2nx)\ {\rm d}x$. 
$(7)$: Sum up $(5)$ and $(6)$. 
$(8),(9)$: Let $z=-1$, $z=i$ in $\displaystyle\sum^\infty_{n=1}\frac{H_n}{n}z^n={\rm Li}_2(z)+\frac{1}{2}\ln^2(1-z)$. 
$(10)$: Apply $(7)$, $(8)$, $(9)$ to $(4)$.
