Summation involving von Mangoldt function

do you mean [tex] \sum _{n=2}^{\infty} \frac{ \Lambda (n) -1}{n} [/tex] ??


i think is divergent

 

[tex]\frac1n \sum_{k=1}^n\Lambda(k)=1+o(1/\log n)[/tex]

so your series seems to be something like

[tex]\sum\frac{1}{n\log n}\approx\log\log n[/tex]

Obviously this is very heuristic here.

 

OK, it diverges.

[tex]\sum_{n=2}^{\infty} \frac{\Lambda(n) -1}{n}=\sum_p\sum_{k=1}^\infty\(\frac{\log p-1}{p}+\frac{\log p-1}{p^2}+\cdots\)=\sum_p\frac{\log p-1}{p-1}[/tex]
and we all know that
[tex]\sum_p\frac1p=+\infty[/tex]

 

My post addressed the case where n runs from 2 to infinity, which diverges.

 

Right, right... yeah, I calculated it for numerator [itex]\Lambda[/itex] first, forgetting about the -1 term, and when I added it back in forgot that part.

But wouldn't that also suggest divergence (in the other direction), since the prime powers are density 0, the reciprocal primes vary as log log n, and the reciprocal integers vary as log n?

Numerical experimentation would be nice here.