Updates:
Status: Fixed
Comment #15 on issue 1092 by [email protected]: limit(sum(1/k, (k, 1,
n))-log(n), n, oo) should do it
http://code.google.com/p/sympy/issues/detail?id=1092
Bisected to
commit 70243d5e83121e0d49ad0970e96adf27d952333a
Author: Raoul Bourquin <[email protected]>
Date: Sat Feb 23 19:16:52 2013 +0100
Make harmonic numbers tractable for Gruntz algorithm.
This adds nothing really new but consistency.
In [11]: gruntz(harmonic(n), n, oo)
Out[11]: oo
In [12]: gruntz(harmonic(n, 2), n, oo)
Out[12]: pi**2/6
In [13]: gruntz(harmonic(n, 3), n, oo)
Out[13]: -polygamma(2, 1)/2
In [14]: gruntz(harmonic(n, 4), n, oo)
Out[14]: pi**4/90
In [15]: gruntz(harmonic(n, 5), n, oo)
Out[15]: -polygamma(4, 1)/24
In [16]: gruntz(harmonic(n, 6), n, oo)
Out[16]: pi**6/945
We can already do this by direct computation:
In [40]: harmonic(oo)
Out[40]: zoo
In [41]: harmonic(oo, 2)
Out[41]: pi**2/6
In [42]: harmonic(oo, 3)
Out[42]: zeta(3)
In [43]: harmonic(oo, 4)
Out[43]: pi**4/90
In [44]: harmonic(oo, 5)
Out[44]: zeta(5)
In [45]: harmonic(oo, 6)
Out[45]: pi**6/945
Note how the polygamma correspond to the zeta functions:
In [14]: gruntz(harmonic(n, 7), n, oo)
Out[14]: -polygamma(6, 1)/720
In [15]: gruntz(harmonic(n, 7), n, oo).rewrite(zeta)
Out[15]: zeta(7)
In [16]: harmonic(oo, 7)
Out[16]: zeta(7)
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