Logarithmic p-Series: MathCad Findings

[notknowing]

Homework Statement

According to textbooks, the logarithmic p-series given by
[tex]\sum_{n=2}^n \frac{1}{n \ ln(n)^p } [\tex] and should converge when p>1 and diverge when [tex]p \leq 1 [\tex]

Homework Equations

Using MathCad (version 11 to 14), I find that the corresponding integral
[tex]int_{2}^{infty} \frac {1}{x \ {ln(x)}^p} dx [\tex] always converges. For instance, for p=0.6, I find that the integral becomes 49.916 (instead of diverging)

The Attempt at a Solution

I have never before encountered a problem with MathCad, so this discrepancy is really surprising. I'm just curious about reactions or observations of similar problems with MathCad.

 

[Gib Z]

I don't use MatCad, however the numerical answer you provide me implies that MathCad integrates numerically (as opposed to symbolically, like Mathematicia). It may happen that MathCad is using a numerical integration technique that isn't effective over infinite regions of integration.

Perhaps we could try an experiment? On MathCad, please integrate 1/x between 1 and infinity, I wouldn't be overly surprised if it gave some finite answer. Both integrands in question, 1/x and the p integral, may appear to have a finite integral, they do have some properties required, such as terms approaching zero.

To see why your corresponding integral actually doesn't always converge, use the fundamental theorem of calculus after a substitution.

 

[notknowing]

I don't use MatCad, however the numerical answer you provide me implies that MathCad integrates numerically (as opposed to symbolically, like Mathematicia). It may happen that MathCad is using a numerical integration technique that isn't effective over infinite regions of integration.

Perhaps we could try an experiment? On MathCad, please integrate 1/x between 1 and infinity, I wouldn't be overly surprised if it gave some finite answer. Both integrands in question, 1/x and the p integral, may appear to have a finite integral, they do have some properties required, such as terms approaching zero.

To see why your corresponding integral actually doesn't always converge, use the fundamental theorem of calculus after a substitution.

Hi, thanks for the reply. I tried the integral you mentioned but this resulted in no answer (not convergent).

 

[Gib Z]

Why does showing that it is not convergent mean there is no answer?