Skip to main content

Timeline for The sum of a series, continued

Current License: CC BY-SA 3.0

8 events
when toggle format what by license comment
Sep 4, 2015 at 21:12 history edited GH from MO CC BY-SA 3.0
added 6 characters in body
Sep 4, 2015 at 21:05 comment added GH from MO I think we need to plug in $x=-1$ instead of $x=-2$, and this works for any $q>1$. This is because your sum on the left hand side equals $\sum_{k=0}^\infty\frac{(-qx)^k}{(q;q)_k}$.
Sep 4, 2015 at 19:29 comment added GH from MO @Deepti: Thank you!
Sep 4, 2015 at 19:10 comment added Deepti @GHfromMO One of the possible references is: G. E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge Univ. Press, Cambridge 1999, Corollary 10.2.2 (b) on page 490. It is also presented in the Books by Kac&Cheung, and Gasper&Rahman, but I don't have the exact pages at the moment (Can provşde later on)
Sep 4, 2015 at 18:58 comment added Gjergji Zaimi @GHfromMO, A quick way to prove it is to notice that the coefficient of $x^kq^{-n}$ on both sides is the number of partitions of $n$ into $k$ distinct parts.
Sep 4, 2015 at 18:53 comment added GH from MO Can you give a precise reference for this Euler identity?
Sep 4, 2015 at 18:46 history edited GH from MO CC BY-SA 3.0
added 10 characters in body
Sep 4, 2015 at 18:19 history answered Deepti CC BY-SA 3.0