Timeline for The sum of a series, continued
Current License: CC BY-SA 3.0
8 events
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Sep 4, 2015 at 21:12 | history | edited | GH from MO | CC BY-SA 3.0 |
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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 |
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Sep 4, 2015 at 18:19 | history | answered | Deepti | CC BY-SA 3.0 |