Timeline for Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

Current License: CC BY-SA 3.0

15 events

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Apr 13, 2017 at 12:20	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Dec 17, 2013 at 11:31	vote	accept	user91500
Dec 17, 2013 at 0:20	comment	added	Raymond Manzoni		@MikeSpivey: here too simpler answers could be given to evaluate directly $\;\displaystyle\sum_{n=1}^\infty \frac{H_{2n}}{n^2}$ or possibly $\;\displaystyle\sum_{n=1}^\infty \frac{H_{2n+1}}{n^2}$ (but these things take much time so...). You were indeed lucky !
Dec 17, 2013 at 0:14	comment	added	Mike Spivey		@RaymondManzoni: You're welcome. It was one of those questions where the quality of the answers far exceeded my expectations!
Dec 17, 2013 at 0:08	comment	added	Raymond Manzoni		@MikeSpivey: really neat trick from robjohn indeed (and an impressive thread!). Many thanks to share Mike,
Dec 16, 2013 at 23:56	comment	added	Mike Spivey		+1. For a different evaluation of $S(-1)$ that relies solely on manipulation of the summation, see robjohn's answer here.
Dec 16, 2013 at 23:29	history	edited	Raymond Manzoni	CC BY-SA 3.0	More details, minor correction
Dec 16, 2013 at 16:49	history	edited	Raymond Manzoni	CC BY-SA 3.0	added 103 characters in body
Dec 16, 2013 at 16:44	comment	added	Raymond Manzoni		You are right @ALGEAN I had two signs wrong since the odd terms had to cancel (corrected).
Dec 16, 2013 at 16:42	history	edited	Raymond Manzoni	CC BY-SA 3.0	two wrong signs...
Dec 16, 2013 at 16:15	comment	added	user91500		If we let $n=2m$ then $S(1)-S(-1)=0$.
Dec 16, 2013 at 16:10	comment	added	user91500		Thanks very much. Why $S(1)-S(-1)=2\sum_{m=1}^\infty \frac{H_{2m}}{(2m)^2}$?
Dec 16, 2013 at 15:53	history	edited	Raymond Manzoni	CC BY-SA 3.0	generalization
Dec 16, 2013 at 15:18	history	edited	Raymond Manzoni	CC BY-SA 3.0	case k=5
Dec 16, 2013 at 14:53	history	answered	Raymond Manzoni	CC BY-SA 3.0