Convergence of \sum_{n=0}^{\infty} zeta^{(n)}(s)

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Homework Help Overview

The discussion centers around the convergence of the series \(\sum_{n=0}^{\infty} zeta^{(n)}(s)\), where \(n\) represents the \(n\)th derivative of the Riemann Zeta function. Participants are exploring whether this sum converges for all values of \(s\) and what it converges to if it does.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • One participant suggests using the integral test and expresses a belief that the series diverges, supported by graphical analysis. Others question the nature of divergence, with some proposing the concept of Borel resummation as a method to handle the series.

Discussion Status

The conversation is ongoing, with participants raising different perspectives on the convergence of the series. There is a mix of attempts to prove divergence and discussions about alternative methods like Borel resummation. No consensus has been reached yet.

Contextual Notes

Participants are working under the assumption that the series may diverge, but they are also exploring the implications of this divergence and potential methods to analyze it further.

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Homework Statement



Does the sum [tex]\sum_{n=0}^{\infty} zeta^{(n)}(s)[/tex] converge regardless of s, where n is the nth derivative of the Riemann Zeta function? If it converges tell what value it converges to.

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The Attempt at a Solution



I used the integral test, and I think it diverges. Plus, by plotting the sequence on Matematica, it looks like it is diverging. Am I correct?
 
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Divergent, but you can still resum the series using Borel resummation.
 


How do i do that?
 


Can anyone prove that the sum [tex]\sum_{n=0}^{\infty} zeta^{(n)}(s)[/tex] goes to INFINITY not just diverges
 


seanhbailey said:
Can anyone prove that the sum [tex]\sum_{n=0}^{\infty} zeta^{(n)}(s)[/tex] goes to INFINITY not just diverges

Apply the operator 1 - d/ds to the summation. To investigate convergence, you can apply it to the partial sums of the first n terms. If you apply it formally to the infinite summation then you get the same result you would get after performing the Borel resummation.
 

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