## prime_n^(1/n) – prime_(n+1)^(1/(n+1))

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• a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) ,  {n,1,k} means n=1 to k. For k 1.   What is the largest
Message 1 of 4 , Feb 2, 2012
a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) ,  {n,1,k} means n=1 to k.
For k <= 1,000,000, observe a < 1, and a-->1.

What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

John W. Nicholson

[Non-text portions of this message have been removed]
• I do not know why the extra characters were place in, but here it is again. a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to
Message 2 of 4 , Feb 2, 2012
I do not know why the extra characters were place in, but here it is again.

a := sum ({n,1,k}, prime_n^(1/n)  prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to k.
For k <= 1,000,000, observe a < 1, and a-->1.

What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

John W. Nicholson

--- In primenumbers@yahoogroups.com, "John W. Nicholson" <reddwarf2956@...> wrote:
>
> a := sum ({n,1,k},Â prime_n^(1/n) â" prime_(n+1)^(1/(n+1)))Â ,Â Â {n,1,k} means n=1 to k.
> For k <= 1,000,000, observe a < 1, and a-->1.
>
> Â
> What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?
> Â
> John W. Nicholson
>
> [Non-text portions of this message have been removed]
>
• ... which was an unhelpful way of defining a(k) = 2 - prime(k+1)^(1/(k+1)) At large k, the PNT gives (1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k)) David
Message 3 of 4 , Feb 2, 2012
"John" <reddwarf2956@...> wrote:

> a := sum ({n,1,k}, prime_n^(1/n)  prime_(n+1)^(1/(n+1)))

which was an unhelpful way of defining

a(k) = 2 - prime(k+1)^(1/(k+1))

At large k, the PNT gives

(1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k))

David
• ... but meant to write (1 - a(k)) = (log(k)/k)*(1 + O(log(log(k))/log(k)))
Message 4 of 4 , Feb 2, 2012