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• ... 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
Message 1 of 3 , Jul 21, 2007
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--- "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.
> How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?
> (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 the tail
from both sides.

Phil

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• ... proven. ... the tail ... Thank you. Remains adding that the convergence radius r of sum(1/a^ln n) and sum(1/a^ln p) is r=a e corresponding to the unity in
Message 1 of 3 , Jul 22, 2007
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--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
wrote:
>
> --- "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.
> > How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge?
> > (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
the tail
> from both sides.
>
> 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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