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Re: SumConvergence mistake?

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Of course, in this case, one needs to know not only that the limit of the sequence of terms tends to zero but also that the sequence of absolute values is monotonically decreasing. But Reduce can prove that:

Reduce[n^(1/n) > (n + 1)^(1/(n + 1)),  n,  Integers]
Element[n,  Integers] && n >= 3

Hence Mathematica ought to be able to apply the Leibniz test in this case also.

Andrzej Kozlowski


On 3 Apr 2011, at 13:58, Andrzej Kozlowski wrote:
>
> But in other, similar cases, it can't:
>
> SumConvergence[(-1)^n (n^(1/n) - 1), n]
>
> SumConvergence[(-1)^n (n^(1/n)-1),n]
>
> even though Limit knows that:
>
> Limit[(n^(1/n) - 1), n -> Infinity]
>
> 0
>
>
> It looks like SumConvergence would not do very well in our first year analysis exam.
>
> Andrzej Kozlowski
>
>
> On 3 Apr 2011, at 12:59, DrMajorBob wrote:
>
>> You are right, and SumConvergence is wrong:
>>
>> NSum[(-1)^k Log[k]/k, {k, Infinity}]
>>
>> 0.159869
>>
>> Sum[(-1)^k Log[k]/k, {k, Infinity}]
>> N@%
>>
>> 1/2 (2 EulerGamma Log[2] - Log[2]^2)
>>
>> 0.159869
>>
>> Bobby
>>
>> On Sat, 02 Apr 2011 17:06:42 -0500, Jon Joseph <josco.jon@gmail.com> wrote:
>>
>>> All: The following test for conditional convergence appears to give the
>>> wrong answer. By the Alternating Series test I think the sum should 
>>> conditionally converge. Any comment or correction?
>>>
>>> SumConvergence[(-1)^k Log[k]/k, k]
>>>
>>> False (* I think should be true *)
>>>
>>> and
>>>
>>> SumConvergence[ Log[k]/k, k]
>>>
>>> is correctly
>>>
>>> False
>>>
>>> Jon.
>>
>>
>> --
>> DrMajorBob@yahoo.com
>>
>

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Re: SumConvergence mistake? Andrzej Kozlowski <akoz@mimuw.edu.pl> - 2011-04-04 10:31 +0000

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