Timeline for iterated harmonic numbers vs Riemann zeta
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2017 at 19:22 | comment | added | GH from MO | See also my response. | |
| Jun 18, 2017 at 19:15 | vote | accept | T. Amdeberhan | ||
| Jun 18, 2017 at 19:09 | comment | added | T. Amdeberhan | @Lucia: I know. Not to worry. You get a cool bonus in the end. :-) | |
| Jun 18, 2017 at 19:07 | comment | added | Lewi_Sol | I appreciate this. | |
| Jun 18, 2017 at 19:06 | comment | added | GH from MO | @Lewi_Sol: (continuing Lucia's comment) ... and each combination contributes $\zeta(m+2)$. | |
| Jun 18, 2017 at 19:05 | comment | added | Lucia | @T. Amdeberhan: I don't mind your edit, but when I answered the question, there was only Question 1! | |
| Jun 18, 2017 at 19:03 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
added 29 characters in body
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| Jun 18, 2017 at 18:54 | comment | added | Lucia | @Lewi_Sol: In expanding out the harmonic sums, all that matters is how many of them are distinct. Since there are $m+1$ terms (including $n$) this gives $m+1$ possible combinations. | |
| Jun 18, 2017 at 18:52 | comment | added | Lewi_Sol | Lucia: great reference. Compiling all the places wherever equality occurs (such as your examples, $\zeta(2,2), \zeta(2,1,1),$ etc requires additional justification, especially in higher dimensions. What do you do there? | |
| Jun 18, 2017 at 18:41 | comment | added | GH from MO | Nice proof. Actually, the OP's left hand side is the multiple zeta star value $\zeta^\ast(2,1_m)$, so chances are that his identity is already in the literature. | |
| Jun 18, 2017 at 18:30 | history | answered | Lucia | CC BY-SA 3.0 |