summation involving von Mangoldt function

[mathslover]

Please help me in solving the problem,
find the sum
Sum{r=2 to infinity} (von Mangoldt(r)-1)/r



Your help is appreciated.

[mhill]

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


i think is divergent

[mathslover]

hi mhill,
can you prove that the series is divergent?



-Ng

[CRGreathouse]

hi mhill,
can you prove that the series is divergent?

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

so your series seems to be something like

\sum\frac{1}{n\log n}\approx\log\log n

Obviously this is very heuristic here.

[CRGreathouse]

OK, it diverges.

\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}
and we all know that
\sum_p\frac1p=+\infty

[mathslover]

when n is a prime or prime power, the summation is okay.

but suppose when n=6, then the sum is (von Mangoldt(6) -1)/6 , which is = -1/6,

as n runs from 2 to infinity,can we settle the problem of convergency or divergency?



-Ng

[CRGreathouse]

when n is a prime or prime power, the summation is okay.

but suppose when n=6, then the sum is (von Mangoldt(6) -1)/6 , which is = -1/6,

as n runs from 2 to infinity,can we settle the problem of convergency or divergency?

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

[mathslover]

Hi CRGreatHouse,

In your post 1673, the summation on LHS runs from n=2 to infinity, (n=2,3,4,5,6,7,8,...)

But the summation on RHS runs over all primes.(p=2,3,5,7,...)


From the definition of von Mangoldt function,when n=6,10,12,14,15,18,... , the summand

became (-1/n) whenever n is not equal to any prime or prime power.

Is something missing ?


-Ng

[CRGreathouse]

Is something missing ?

Right, right... yeah, I calculated it for numerator \Lambda 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.

[mathslover]

I have tried numerical calculation and the sum seems to converge to ~ -1.16

Can we approach the problem from Zeta function?



-Ng