I was working on a problem recently which went like this...prove:
$$\sum\limits_{n\geq 1}\binom{3n}{2n}8^{-n}=\frac{3}{\sqrt{5}}$$
I was able to determine that
$$\mathcal{G}(x)=\sum\limits_{n\geq 1}\binom{3n}{2n}x^{-n}= \frac{\sqrt{3}}{2\pi}\Bigg[\mathcal{I}\left(\frac{1}{3},\frac{3}{2}\sqrt{3x}\right)+\mathcal{I}\left(\frac{1}{3},-\frac{3}{2}\sqrt{3x}\right)\Bigg]$$
whereby
$$\mathcal{I}(z,\alpha)=\frac{-\pi}{\sin(\pi z)}\mathcal{U}_{-z-1}(\alpha)$$
and $\mathcal{U}_{z}(\alpha)$ are Chebyshev's Polynomials of the second kind. Alternativley, Wolfram Alpha gives me a second closed-form expression for the infinite sum...namely
$$\mathcal{G}(x)=2\cos\Bigg[\frac{1}{3}\arcsin\left(\frac{3\sqrt{3x}}{2}\right)\Bigg](4-27x)^{-1/2}$$
A friend had worked out that $\mathcal{G}(x)$ is a solution to the following differential equation:
$$\left(27x^{2}-4x\right)\mathcal{G}^{\prime\prime}+(54x-2)\mathcal{G}^{\prime}+6\mathcal{G}=0,\quad \mathcal{G}(0)=1,\mathcal{G}^{\prime}(0)=3$$
I'm wondering about how we would come to solve this differential equation because it took me a lot of work to derive my solution in terms of Chebyshev polynomials and thought maybe the differential equation route would be simpler. I understand this is a Sturm-Louiville problem but I lack the mathematical knowledge to apply it to the differential. Any help would be greatly appreciated!
Considering the differential equation $$x\left(27x-4\right)G''+2(27x-1)G'+6{G}=0$$ let $G=\frac H {\sqrt{4-27x}}$ to make $$4 x (27 x-4) H''(x)+4 (27 x-2) H'(x)-3 H(x)=0$$ the solution of which being $$H=c_1 \sinh \left(\frac{1}{3} \log \left(9 \sqrt{x}+\sqrt{81 x-12}\right)\right)+c_2 \cosh \left(\frac{1}{3} \log \left(9 \sqrt{x}+\sqrt{81 x-12}\right)\right)$$ Back to $G$ and using the conditions, after simplifications, $$G=\frac{2 }{\sqrt{4-27 x}}\cos \left(\frac{1}{3} \sec ^{-1}\left(\frac{2}{\sqrt{4-27 x}}\right)\right)$$ which, numerically, is "close" to your expression but not the same.
Expanding the result as Taylor series, the above write $$G=1+3 x+15 x^2+84 x^3+495 x^4+O\left(x^{5}\right)$$ while, for your result, $$1+\frac{27 }{8}x+\frac{2139 }{128}x^2+\frac{97119 }{1024}x^3+\frac{18363603 }{32768}x^4+O\left(x^5\right)$$ Not far but not the same. May be a typo in the differential equation.