Let $\varphi(n):=(-1)^{n+1}(n+1)2^{2n}$.


I am able to prove the following identity (${\color{red}{\mathbf{LHS}}}$=infinite series, ${\color{blue}{\mathbf{RHS}}}$=finite sum)
\begin{align\*}
{\color{red}{\frac{1}{\varphi(n)}\sum\_{k\geq1}\frac{(-1)^k}{\binom{3n+k+2}{2n+2}}}}
&={\color{blue}{\sum\_{j=1}^n\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}+3-4\ln 2}}. \tag1
\end{align\*}
Replacing $n=0$ in equation (1) leads to $\displaystyle\sum\_{j\geq1}\frac{(-1)^j}{\binom{j+2}2}=4\ln 2-3$.


Letting $n\rightarrow\infty$ in (1) unearths the less trivial result
$$\sum\_{j=1}^{\infty}\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}=4\ln 2-3.$$



> 
> **QUESTION.** How can one exercise an analytic continuation of the ${\color{blue}{\mathbf{RHS}}}$ in (1), ${\color{purple}{\text{in the variable $n$}}}$, so that we would be able to compute at $n=-\frac23$ and derive (from it) a value for the infinite series on the ${\color{red}{\mathbf{LHS}}}$, i.e.
> $$\frac1{\varphi(-\frac23)}\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}\,?$$
> 
> 
> 



A need for analytic continuation of a finite sum function
As for the sum $$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}$$
one can evaluate it by means of the Beta function integral, like in this recent [computation](https://mathoverflow.net/questions/449776/closed-form-of-an-infinite-series/449810#449810).
$$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}=\frac23\sum\_{k\geq1}(-1)^k \frac{\Gamma\big(k+\frac13\Big)\Gamma\big(\frac23\big)}{\Gamma(k+1)}=\frac23\sum\_{k\geq1}(-1)^kB\Big(k+\frac13,\frac23\Big)=$$
$$=\frac23\sum\_{k\geq1} \int\_0^1(-1)^k {x^{k-\frac23}}{(1-x)^{-\frac13}}dx=-\frac23 \int\_0^1\frac{1}{1+x} \Big(\frac{x }{1-x}\Big)^{1/3}dx=$$


The latter integral is easily rationalised by changing variable; we get $$=-\int\_0^\infty\frac{2t^3}{(t^3+1)(2t^3+1)}dt=\int\_0^\infty\frac{2dt}{2t^3+1 } -\int\_0^\infty\frac{2dt}{ t^3+1}=$$$$=(\sqrt[3]4-2)\int\_0^\infty\frac{dt}{ t^3+1}$$ whence $$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}= 
\frac{2\pi}{3\sqrt3} (\sqrt[3]4-2)$$
which evaluates to $-0.4989..$.


