Timeline for How do I evaluate this sum :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n} $?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 14, 2016 at 20:01 | vote | accept | CommunityBot | ||
| Oct 14, 2016 at 2:25 | answer | added | T. Amdeberhan | timeline score: 7 | |
| S Oct 13, 2016 at 22:38 | history | suggested | Martin Sleziak |
Removed (harmonic-functions) tag
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| Oct 13, 2016 at 22:37 | review | Suggested edits | |||
| S Oct 13, 2016 at 22:38 | |||||
| Oct 13, 2016 at 21:53 | answer | added | Julian Rosen | timeline score: 19 | |
| Oct 13, 2016 at 21:47 | comment | added | user99666 | i meant sum(1/n^k) , for k greater then 1 | |
| Oct 13, 2016 at 21:46 | comment | added | Gerry Myerson | What is "the Riemann series"? | |
| Oct 13, 2016 at 20:22 | comment | added | Carlo Beenakker | with the $2^n$ in the exponent it's 0.96630012055722... do you really need more accuracy? | |
| Oct 13, 2016 at 20:08 | comment | added | user99666 | sorry , I have a wrong typo I meant :$\sum_{n=1}^{\infty}\frac{H_{n}^3}{(n+1)2^n} $ not $3^n$ in the denominator and i edited the question | |
| Oct 13, 2016 at 20:08 | history | edited | user99666 | CC BY-SA 3.0 |
added 66 characters in body; edited title
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| Oct 13, 2016 at 19:58 | comment | added | Carlo Beenakker | it evaluates to 0.38360530295885199... --- which is not recognized as having a closed form representation by the Inverse Symbolic Calculator | |
| S Oct 13, 2016 at 19:50 | history | suggested | T. Amdeberhan | CC BY-SA 3.0 |
clarity added, reformulation considered.
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| Oct 13, 2016 at 19:49 | review | Suggested edits | |||
| S Oct 13, 2016 at 19:50 | |||||
| Oct 13, 2016 at 19:42 | review | First posts | |||
| Oct 13, 2016 at 19:52 | |||||
| Oct 13, 2016 at 19:36 | history | asked | user99666 | CC BY-SA 3.0 |