Timeline for Series expansion with remaining $log n$
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| May 17, 2013 at 12:47 | comment | added | Carlo Beenakker | the series expansion is easiest if you first take the logarithm, and then you find directly a powerseries in $n^{-1}\ln n$, $$-\frac{k(k+1)}{2n}\ln n+2\ln(2kn)\left[\sum_{p=0}^{\infty}\frac{1}{p}k^p(1+k)^p(2kn)^{-p}(\ln n)^p \right]$$ | |
| May 17, 2013 at 9:19 | comment | added | ELW | It seems that it is the $n^{k(k +1)/(2n)}$? | |
| May 17, 2013 at 9:00 | history | asked | ELW | CC BY-SA 3.0 |