Sum of the sum of harmonic series?

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Homework Help Overview

The discussion revolves around the convergence or divergence of the series Ʃ1/(1+2+3+4+5...+n) as n approaches infinity. Participants are exploring the implications of rewriting the series and comparing it to known divergent series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to rewrite the series and question whether the transformation to Ʃ(Ʃ1/n) is valid. There is discussion about the relationship between the series and the known divergence of Ʃ(1/n). Some participants also explore the implications of rewriting the sum of integers from 1 to n.

Discussion Status

The discussion is active, with participants questioning their assumptions and exploring different interpretations of the series. Some guidance has been offered regarding the formula for the sum of integers, which has prompted further reflection on the original problem.

Contextual Notes

There is a focus on the mathematical properties of series and the need to clarify the assumptions underlying the convergence or divergence of the given series. Participants are also considering the implications of rewriting the series in different forms.

Nikitin
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Homework Statement


Does this converge or diverge?

Ʃ1/(1+2+3+4+5...+n), as n---> infinity?

The Attempt at a Solution



I rewrote this into Ʃ(Ʃ1/n) (is it correct?).

I figured that since Ʃ(1/n) diverges, then the sum of each partial sum most (obviously) also diverge.

However, it appears I'm mistaken. Can somebody help?
 
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Nikitin said:

Homework Statement


Does this converge or diverge?

Ʃ1/(1+2+3+4+5...+n), as n---> infinity?


The Attempt at a Solution



I rewrote this into Ʃ(Ʃ1/n) (is it correct?).

I figured that since Ʃ(1/n) diverges, then the sum of each partial sum most (obviously) also diverge.

However, it appears I'm mistaken. Can somebody help?

Is 1/(1+2) equal to (1/1) + (1/2)? Basically, you are claiming that the answer is yes.

RGV
 
oh crap. yeh you're right.
 
1+2+...+n=n(n+1)/2,so compare with the convergent series 2/n(n+1)
 
ahh thanks. i completely forgot that you could rewrite 1+2+3..+n into n(n+1)/2.. thx!
 

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