Timeline for Showing an alternating integer series is never $0$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 14, 2018 at 17:42 | comment | added | Max Alekseyev | It may be worth to notice that the sum equals the coefficient of $z^{2m-1}$ in $$\frac{(2m+2)!(2m-1)!}{(m+1)!} e^{2m(m+1)z}\int_0^1 (1-e^{-(2m+1)z}-u)^mu^{m+1}du.$$ | |
| Mar 13, 2018 at 20:27 | answer | added | skbmoore | timeline score: 4 | |
| Mar 8, 2018 at 1:56 | comment | added | Pietro Majer | It may be convenient to see the sum as the $(2m+2)$-th iterated backwards finite difference, with spacing $h=2m+1$, of the function $f(x):=x_+^{2m-1}$, computed at the point $2m(m+1)$. | |
| Mar 8, 2018 at 0:08 | comment | added | Pietro Majer | Dividing by $-(2m+1)^{2m-1}$ one is left with a sum which is very similar to $a\big({2m(m+1)\over2m+1},2m-1,2m+2\big)$ for the numbers $a(p,n,m)$ defined here: mathoverflow.net/questions/29118/… --It may be a start. | |
| Mar 7, 2018 at 22:55 | history | edited | Bill Trok | CC BY-SA 3.0 |
edited body
|
| Mar 7, 2018 at 22:55 | comment | added | Bill Trok | @MaxAlekseyev You are correct. I suppose I meant m > 1. I'll edit the post. | |
| Mar 7, 2018 at 22:00 | comment | added | Max Alekseyev | The sum is 0 for $m=1$. | |
| Mar 7, 2018 at 3:54 | review | First posts | |||
| Mar 7, 2018 at 4:08 | |||||
| Mar 7, 2018 at 3:50 | history | asked | Bill Trok | CC BY-SA 3.0 |