- --- "Werner D. Sand" <Theo.3.1415@...> wrote:
> The divergence of sum(1/2^ln n) and sum(1/2^ln p) can easily be proven.

3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3)

> How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?

> (IF they do. What are the limits?)

sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of

reciprocals converges.

=> sum(1/p^ln(3)) also converges.

The numerical value? Sum the start and integrate functions bounding the tail

from both sides.

Phil

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http://answers.yahoo.com/dir/?link=list&sid=396545469 - --- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>

wrote:>

proven.

> --- "Werner D. Sand" <Theo.3.1415@...> wrote:

> > The divergence of sum(1/2^ln n) and sum(1/2^ln p) can easily be

> > How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?

the tail

> > (IF they do. What are the limits?)

>

> 3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3)

> sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of

> reciprocals converges.

>

> => sum(1/p^ln(3)) also converges.

>

> The numerical value? Sum the start and integrate functions bounding

> from both sides.

Thank you. Remains adding that the convergence radius r of sum(1/a^ln

>

> Phil

>

n) and sum(1/a^ln p) is r=a>e corresponding to the unity in Riemann's

Zeta function.

Werner

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