On sums such as $\sum_{k=0}^\infty \binom{2k}{k}\frac{1}{8^k}=\sqrt{2}$

The Lagrange Inversion Formula provides an appropriate method to derive \begin{align*} \sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}=\frac{4\sqrt{10}}{5}\cos\Big(\frac{1}{3}\arcsin \frac{3\sqrt{6}}{8}\Big)\tag{1} \end{align*}.

Let a formal power series $w=w(t)$ be implicitely defined by a relation $w=t\Phi(w)$, where $\Phi(t)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:

$$[t^n]w(t)^k=\frac{k}{n}[t^{n-k}]\Phi(t)^n$$

A variation stated as formula $G6$ in Lagrange Inversion: when and how by R. Sprugnoli (etal) is:

Let $F(t)$ be any formal power series and $w=t\Phi(w)$ as before, then the following is valid:

\begin{align*} [t^n]F(t)\Phi(t)^n=\left[\left.\frac{F(w)}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\tag{2} \end{align*}

Note: The notation $[\left.f(w)\right|w=g(t)]$ is a linearization of $\left.f(w)\right|_{w=g(t)}$ and denotes the substitution of $g(t)$ to every occurrence of $w$ in $f(w)$ (that is, $f(g(t))$). In particular, $w=t\Phi(w)$ is to be solved in $w=w(t)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.

In order to prove (1) we set $F(t)=1$ and $\Phi(t)=(1+t)^3$. We then have

$$t\Phi'(w)=3t(1+w)^2=\frac{3t\Phi(w)}{1+w}=\frac{3w}{1+w}$$

It follows:

\begin{align*} \binom{3n}{n}&=[t^n]F(t)\Phi(t)^n=[t^n](1+t)^{3n}\\ &=[t^n]\left[\left.\frac{1}{1-t\Phi'(w)}\right|w=t\Phi(w)\right]\\ &=[t^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=t\Phi(w)\right]\\ &=[t^n]\left[\left.\frac{1+w}{1-2w}\right|w=t\Phi(w)\right]\\ \end{align*} Let \begin{align*} A(t):=\sum_{n\ge0}\binom{3n}{n}t^n=\left.\frac{1+w}{1-2w}\right|_{w=t\Phi(w)} \end{align*}

Expressing $A(t)=\frac{1+w}{1-2w}$ in terms of $w$, we get

$$w=\frac{A(t)-1}{2A(t)+1}$$

Since $w=t\Phi(w)=t(1+w)^3$, we obtain \begin{align*} \frac{A(t)-1}{2A(t)+1}=t\left(1+\frac{A(t)-1}{2A(t)+1}\right)^3 \end{align*}

which simplifies to: \begin{align*} (4-27t)A(t)^3-3A(t)-1=0\tag{3} \end{align*}

In order to get the RHS of $(1)$ we first analyse the structure of (3) which is

$$f(t)A(t)^3-3A(t)=1$$

with $f(t)$ linear and observe a similarity of this structure with the identity

$$4\cos^3{t}-3\cos{t}=\cos{3t}$$

Thus we use the ansatz:

\begin{align*} A(t) := \frac{2\cos\left(g(t)\right)}{\sqrt{4-27t}}\tag{4} \end{align*}

We see

\begin{align*} (4-27t)A(t)^3-3A(t) &=\frac{8\cos^3\left(g(t)\right)}{\sqrt{4-27t}}-\frac{6\cos\left(g(t)\right)}{\sqrt{4-27t}}=\\ &=\frac{2\cos\left(3g(t)\right)}{\sqrt{4-27t}}\\ &=1 \end{align*}

Now, since

\begin{align*} 2\cos\left(3g(t)\right)&=\sqrt{4-27t}\\ 4\cos^2\left(3g(t)\right)&=4-27t\\ \sin^2\left(3g(t)\right)&=\frac{27}{4}t\\ \end{align*}

we get \begin{align*} g(t)&=\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\tag{5}\\ \end{align*}

We finally conclude from (4) and (5) \begin{align*} \color{blue}{\sum_{k=0}^\infty \binom{3k}{k}\frac{1}{8^k}} &=\left.\frac{2\cos(g(t))}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\ &=\left.\frac{2\cos\left(\frac{1}{3}\arcsin\left(\frac{3\sqrt{3t}}{2}\right)\right)}{\sqrt{4-27t}}\right|_{t=\frac{1}{8}}\\ &\,\,\color{blue}{=\frac{4\sqrt{10}}{5}\cos\left(\frac{1}{3}\arcsin\left( \frac{3\sqrt{6}}{8}\right)\right)} \end{align*} and the claim follows.

The binomial theorem can famously prove such sufficiently-small-$x\ge0$ results as$$\sum_{k\ge0}\binom{2k}{k}x^k=(1-4x)^{-1/2}$$(which is useful in deriving the $n$th Catalan number from their generating function) and$$\sum_{k\ge0}\binom{4k}{2k}x^k=\frac{(1-4\sqrt{x})^{-1/2}+(1+4\sqrt{x})^{-1/2}}{\sqrt{2}}$$(this is just the even-$k$ terms of the first result, i.e. the function's even part, after we replace $x$ with $\sqrt{x}$).

I had to resort to WA for$$\sum_{k\ge0}\binom{3k}{k}x^k=\left(1-\frac{27}{4}x\right)^{-1/2}\cos\left(\frac13\arcsin\sqrt{\frac{27x}{4}}\right),$$which can be rewritten with more complicated radicals using$$c:=\cos\left(\frac13\arcsin\sqrt{\frac{27x}{4}}\right)\implies4c^3-3c=1-\frac{27x}{4}.$$If you get such a form using Cardano's method (but I should mention that kinda misses the point), you can use the binomial theorem to prove it. This illustrates the benefits of a hybrid use of EESs: get the answer from them, then prove it with some inspiration.

But $\sum_{k\ge0}\binom{5k}{k}x^k$ didn't, in WA's opinion, succumb to the same analysis, which is why you'd need the suggestion of @LordSharktheUnknown's first comment to get the hypergeometric result (the last argument generalises to $\frac{5^5x}{4^4}$).

Another way is to use the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{|z|=1}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz.$$ For example we have $$\sum_{k\geq0}\frac{1}{8^{k}}\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{2k}}{z^{k+1}}dz=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{1}{z}\sum_{k\geq0}\frac{1}{8^{k}}\frac{\left(1+z\right)^{2k}}{z^{k}}dz$$ $$=-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{8}{z^{2}-6z+1}dz$$ and since we have a pole at $z=3-2\sqrt{2}$ we get $$\sum_{k\geq0}\frac{1}{8^{k}}\dbinom{2k}{k}=\color{red}{\sqrt{2}}.$$ In a similar way $$\sum_{k\geq0}\frac{1}{8^{k}}\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{3k}}{z^{k+1}}dz=-\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{8}{z^{3}+3z^{2}-5z+1}dz$$ and since we have a pole at $z=\sqrt{5}-2$ we get $$\sum_{k\geq0}\frac{1}{8^{k}}\dbinom{3k}{k}=\color{red}{1+\frac{3}{\sqrt{5}}}.$$