I encountered this expression generated by mathematica as a sub-step in a problem I am solving. $$\sum\limits_{M=2}^{\infty}\sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}\pi \csc(n \pi) \Gamma(M)}{(n+1)^2 \Gamma(M-n)\Gamma(n+1)}$$
Question: Is there a closed form for this expression in terms of $\log$, $Li_{2}()$, $Li_{3}()$ and similar?
What have I done so far? Not that it is useful, I got it simplified to the following:
$$\sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} a! (M-1)!}{2^{M} 3^{a+1} (M+a+1)^2 (M+a)!}$$ or $$\sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{M 2^{M} \binom{M+a}{a} 3^{a+1} (M+a+1)^2}$$. I don't know how to proceed. A related question I posted earlier is here.
Edit1: I did one more simplification (if this helps anyone to complete the steps needed):
As per this $$\frac{1}{\binom{M+a}{a}} = (M+a+1) \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$
Therefore the expression becomes
$$\sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{M 2^{M} 3^{a+1} (M+a+1)^2} (M+a+1) \int\limits_{0}^{1} t^{a} (1-t)^{M} dt $$
$$= \sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{M (M+a+1) 2^{M} 3^{a+1}} \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$
Taking the partial fractions of $\frac{1}{M(M+a+1)} = \frac{1}{a+1}\left(\frac{1}{M}-\frac{1}{M+a+1}\right)$,
$$= \sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{M (a+1) 2^{M} 3^{a+1}} \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$ $$- \sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{(M+a+1) (a+1) 2^{M} 3^{a+1}} \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$
$$= \int\limits_{0}^{1} \sum\limits_{M=2}^{\infty}\frac{(-1)^{M+1} (1-t)^{M}}{M 2^{M} } \sum\limits_{a=0}^{\infty} \frac{1 }{ (a+1) 3^{a+1}} t^{a} dt$$ $$- \sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{(M+a+1) (a+1) 2^{M} 3^{a+1}} \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$
$$= K - \sum\limits_{M=2}^{\infty}\sum\limits_{a=0}^{\infty} \frac{(-1)^{M+1} }{(M+a+1) (a+1) 2^{M} 3^{a+1}} \int\limits_{0}^{1} t^{a} (1-t)^{M} dt$$
Where K can be calculated by Mathematica in terms of polylog (which is all I am asking for this question/bounty)
I don't know how to simplify the second integral yet. Integral gurus, please help.