Conjecture: $\sum_{n=0}^{\infty}\frac{k^n}{{2n \choose n}}\cdot n^g=A_g+B_g\pi$ This question came from this $page$.
Given: $$\sum_{n=0}^{\infty}\frac{k^n}{{2n \choose n}}\cdot n^g=A_g+B_g\pi=F(g)\tag1$$
where $g\ge0$ are whole numbers and $k=1,2$ and $3$
Conjecture:$$\lim_{g \to \infty}\frac{A_g}{B_g}=\frac{4\pi}{2^{k}}\tag2$$
Examples: $k=1$ and $g=4$
$$f(4)=A_4+B_4\pi=\frac{32}{3}+\frac{238}{81\sqrt{3}}\pi$$  then $$\frac{A_4}{b_4}\approx2\pi,$$ $$\frac{32}{3}\div\frac{238}{81\sqrt{3}}\approx6.2877...(2\pi=6.2831...)$$ 
Examples: $k=2$ and $g=4$
$$f(4)=A_4+B_4\pi=355+113\pi$$  then $$\frac{A_4}{b_4}\approx \pi,$$ $$355\div113\approx3.14159292...(\pi=3.141592654)$$ 
Examples: $k=3$ and $g=4$
$$f(4)=A_4+B_4\pi=29496+\frac{32524}{\sqrt{3}}\pi$$  then $$\frac{A_4}{b_4}\approx \frac{\pi}{2},$$ $$29496\div\frac{32524}{\sqrt{3}}\approx1.57079...(\frac{\pi}{2}=1.570796327...)$$ 
Does this conjecture $(2)$ hold?
 A: This is not a solution, but maybe a start.
In the case $g=0$, Maple says that for $0 \le k < 4$,
$$ \sum_{n=0}^\infty \frac{k^n}{{2n \choose n}} = \frac{4}{4-k} + \frac{4 \sqrt{k}}{(4-k)^{3/2}} \arcsin(\sqrt{k}/2) $$
Call this $G_0(k)$.  Note that $\arcsin(\sqrt{k}/2) = \pi/6$, $\pi/4$, $\pi/3$ for $k=1,2,3$, so that's where we're going to get the $\pi$'s.  
Now since $k^n n^g = \left(k \dfrac{\partial}{\partial k}\right)^g k^n$
your sum
$$ G_g(k) = \sum_{n=0}^\infty \frac{k^n n^g}{{2n \choose n}} = \left( k \frac{\partial}{\partial k}\right)^g G_0(k) $$
This has exponential generating function
$$ \sum_{g=0}^\infty \frac{G_g(k)}{g!} z^g  = G_0(k e^z)$$
Thus the case $g=4$ is $4!$ times the coefficient of $z^4$ in the Maclaurin series of $G_0(k e^z)$, namely
$$
4\,{\frac {\sqrt {k} \left( {k}^{4}+74\,{k}^{3}+516\,{k}^{2}+464\,k+16
 \right) \arcsin \left( \sqrt {k}/2 \right) }{ \left( 4-k \right) 
^{11/2}}}+{\frac {12\,{k}^{4}+460\,{k}^{3}+1624\,{k}^{2}+496\,k}{
 \left( 4-k \right) ^{5}}}
$$
In general it looks like we'll have 
$$G_g(k) = \frac{P_g(k)}{(4-k)^{g+1}} + \frac{\sqrt{k} Q_g(k)}{(4-k)^{g+3/2}} \arcsin\left(\sqrt{k}/2\right) $$
for some polynomials $P_g$ and $Q_g$ of degree $g$.  These satisfy recursions
$$ \eqalign{P_{g+1} &= (4-k) k P'_g + (1+g)k P_g + \frac{k Q_g}{2}\cr
Q_{g+1} &=  (4-k)k Q'_g + (2 + k + g k) Q_g\cr}$$
