Timeline for How can I simplify this sum any further?

Current License: CC BY-SA 4.0

14 events

when toggle format	what		by	license	comment
Nov 17, 2019 at 18:20	history	edited	RobPratt	CC BY-SA 4.0	added 546 characters in body
Nov 17, 2019 at 17:46	history	edited	RobPratt	CC BY-SA 4.0	added 982 characters in body
Nov 17, 2019 at 17:26	comment	added	Susp1cious		@J.M.isnotamathematician My bad, I didn't realize that the formula doesn't substitute every $a$, only the one's which are a coefficient of $k$
Nov 17, 2019 at 15:57	history	edited	RobPratt	CC BY-SA 4.0	edited body
Nov 17, 2019 at 15:48	history	edited	RobPratt	CC BY-SA 4.0	added 55 characters in body
Nov 17, 2019 at 15:20	comment	added	RobPratt		No, I made a different error, which I should have noticed because evaluating at $x=0$ should yield 1. I’ll fix it.
Nov 17, 2019 at 15:08	comment	added	J. M. isn't a mathematician		@Susp1cious, why is the bottom of your binomial coefficient $\ell$ when it is independent of $k$?
Nov 17, 2019 at 9:02	vote	accept	Susp1cious
Nov 17, 2019 at 8:57	comment	added	Susp1cious		My result for $a=2$ is slightly different. Is it possible that you forgot to eliminate the $2$ from the Binomial? Shouldn't the expression be $\sum_{k=0}^\infty b_{2k}=\sum_{k=0}^\infty\biggl(\frac{1}{2}\sum_{j=0}^1\exp(\pi\ ijk)\biggr)\frac{x^k}{k!}\sum_{l\geq0}\binom{k}{l}$
Nov 17, 2019 at 8:17	vote	accept	Susp1cious
Nov 17, 2019 at 9:02
Nov 16, 2019 at 23:26	comment	added	RobPratt		Wilf's generatingfunctionology and Graham, Knuth, et al.'s Concrete Mathematics are good sources for manipulating sums.
Nov 16, 2019 at 22:06	comment	added	Susp1cious		Thank you very much for your answer! Is there an online platform where I can find identities such as the one you mentioned? Or is it just something one has to know from experience?
Nov 16, 2019 at 21:49	history	edited	RobPratt	CC BY-SA 4.0	deleted 6 characters in body
Nov 16, 2019 at 21:42	history	answered	RobPratt	CC BY-SA 4.0