Re: SumConvergence mistake?
- To: mathgroup at smc.vnet.net
- Subject: [mg117909] Re: SumConvergence mistake?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 4 Apr 2011 06:29:16 -0400 (EDT)
It seems that SumConvergence is misusing the Integral Test here:
SumConvergence[(-1)^k Log[k]/k, k, Method -> "IntegralTest"]
False
But the integral test should not be used as it applies only to non-negative monotone decreasing functions. In any case, the same thing seems to happen to many other series that converge by the Leibniz test:
SumConvergence[ (-1)^k 1/Sqrt[k], k,
Method -> "IntegralTest"]
False
However, Sum gives the right answer in this case too:
Sum[(-1)^k*(1/Sqrt[k]), {k, 1, Infinity}]
(Sqrt[2] - 1)*Zeta[1/2]
In some cases SumConvergence appears to use the Leibniz test quite succesfully, e.g.
In[136]:= SumConvergence[ (2^(1/n) - 1), n]
Out[136]= False
In[137]:= SumConvergence[(-1)^n (2^(1/n) - 1), n]
Out[137]= True
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 at 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 at yahoo.com
>