Re: Incorrect Summation
- To: mathgroup at smc.vnet.net
- Subject: [mg129205] Re: Incorrect Summation
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Fri, 21 Dec 2012 05:37:50 -0500 (EST)
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Am 20.12.2012 09:21, schrieb amzoti:
> On Wednesday, December 19, 2012 1:56:26 AM UTC-8, Murray Eisenberg wrote:
>> What did you type, exactly, for Wolfram|Alpha? I tried several "obvious" things, e.g.,
>>
>> sum cos(log(log(k)))/log(k) for k from 2 to infinity
>>
>> but WA was unable even to interpret it correct as asking for the sum of an infinite series.
>>
>> On Dec 18, 2012, at 2:35 AM, amzoti <amzoti at gmail.com> wrote:
>>
>>> Hi All,
>>
>>> can someone try this and see if it returns an incorrect result?
>>
>>> Sum[Cos[Log[Log[k]]]/Log[k],{k,2,Infinity}]
>>
>>> I tried it on WA and believe the result is incorrect.
> Just copy and paste the expression as-is:
> Sum[Cos[Log[Log[k]]]/Log[k],{k,2,Infinity}]
>
> For this example, if you increase the number of points to max on WA, you can see where I think the result is going astray and why this is returning an incorrect result.
>
> Aside:
> It is odd that sometimes WA accepts the user inputs in standard notation, but other times, it seems to get confused.
>
> I would think that if a user knows how to type Mathematica standard syntax correctly, that WA should be able to accept it.
The sum is not absolutely convergent. For series with product terms
there are criteria for convergency, eg Abel or Dirichlet. Both are
failing with this series.
Since Log@Log@k is going very slowly through the positive and negative
intervalls (0,pi/2) (pi/2,3pi/2) of Cos the partial sums of the
divergent series sum_k 1/log(k) pile up more and more.
Wolfram alpha is giving the same answers as Mathematica.
The partial sums seem to go to some -20000 as long as
loglog k is <3pi/2 and alle the cos are negative.
Since at 3pi/2 cos has a zero, a numerical check routine gets a
seemingly constant negative result.
To reach again positive values one had to double the argument of cos:
Solve[Log@Log@k == 3 \[Pi]/2, k] // N
{{k -> 2.2115508823699243`*^48}}
Solve[Log@Log@k == 5 \[Pi]/2, k] // N
{{k -> 5.367502036729873203843743090888014690659`12.530835720713577*^\
1118}}
--
Roland Franzius
- References:
- Incorrect Summation
- From: amzoti <amzoti@gmail.com>
- Incorrect Summation