Timeline for Closed form for $\sum\limits_{M=2}^{\infty}\sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}\pi \csc(n \pi) \Gamma(M)}{(n+1)^2 \Gamma(M-n)\Gamma(n+1)}$

15 events

when toggle format	what		by	license	comment
S Jun 1 at 17:31	history	bounty ended	Srini
S Jun 1 at 17:31	history	notice removed	Srini
Jun 1 at 17:31	vote	accept	Srini
Jun 1 at 3:44	answer	added	David H		timeline score: 4
May 31 at 14:41	history	edited	Srini		Added another tag
May 30 at 21:15	history	edited	Srini	CC BY-SA 4.0	added 1686 characters in body
May 30 at 5:24	comment	added	Nolord		Perhaps, you might want to try partial fraction decomposition on the inverse binomial coefficient, as it is well known (see the wikipedia of the binomial coefficient).
May 30 at 5:00	comment	added	Srini		Thanks Arjun, you are right, I fixed the error
May 30 at 4:59	history	edited	Srini	CC BY-SA 4.0	added 2 characters in body
May 30 at 4:08	comment	added	Arjun Vyavaharkar		The second summation is incorrect. The summands should be of the form $\frac{{(-1)}^{M+1}}{M2^M\color{blue}{M+a\choose a}\color{black}3^{a+1}{(M+a+1)}^2}$.
S May 28 at 21:25	history	bounty started	Srini
S May 28 at 21:25	history	notice added	Srini		Draw attention
May 26 at 21:52	comment	added	Srini		Yes, in my simplification, I took the limit of these things canceling out
May 26 at 19:28	comment	added	GEdgar		So in $\Gamma(M-n)$ we have $M-n \le 0$, so $\Gamma(M-n)$ is undefined ? And in $\cos(n\pi)$ the csc is undefined? Or perhaps you somehow depend on these two canceling each other?
May 26 at 18:16	history	asked	Srini	CC BY-SA 4.0