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MathGroup Archive 2004

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Re: Wrong Limit

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47790] Re: [mg47775] Wrong Limit
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 27 Apr 2004 04:46:33 -0400 (EDT)
  • References: <200404260641.CAA06357@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

First of all, Mathematica 5.0 gives:


Sum[1/k^p, {k, 1, Infinity}]


Zeta[p]

which is correct so there is no real need to take any limits. Still, 
your observation seems to reveal a bug in Mathematica 5.0. which may be 
related to one that was already discussed here recently.


First, look at this:


Sum[1/k^3,{k,1,m}]


HarmonicNumber[m, 3] + PolyGamma[2, 1]/2 + Zeta[3]

This looks strange, but in fact is correct since:


FullSimplify[PolyGamma[2,1]/2+Zeta[3]]

0

Now


FullSimplify[Sum[1/k^3,{k,1,m}]]


PolyGamma[2, 1 + m]/2 + Zeta[3]


This is still correct:


PolyGamma[2,1+m]/2==HarmonicNumber[m,3]+PolyGamma[2,1]/2//FullSimplify


True

However, it tells us that Limit[PolyGamma[2, 1 + m]/2,m->Infinity] 
ought to be zero, while Mathematica gives:


Limit[PolyGamma[2,1+m]/2,m->Infinity]

Infinity


Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/




On 26 Apr 2004, at 15:41, Ray wrote:

> In Mathematica 4.2, if s[n_]:= Sum[1/k^p,{k,1,n}], then the output for
> Limit[s[n],n->Infinity] was Limit[HarmonicNumber[n,p],n->Infinity].
> Under 5.0.1, the answer to Limit[s[n],n->Infinity] is given incorrectly
> as Infinity for odd p and the actual numerical value for even p. Anyone
> know why Mathematica now gives an incorrect result here for odd p. 
> Thanks.
>
>
>


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