## summation involving von Mangoldt function

find the sum
Sum{r=2 to infinity} (von Mangoldt(r)-1)/r

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 do you mean $$\sum _{n=2}^{\infty} \frac{ \Lambda (n) -1}{n}$$ ?? i think is divergent
 hi mhill, can you prove that the series is divergent? -Ng

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## summation involving von Mangoldt function

 Quote by mathslover 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.
 Recognitions: Homework Help Science Advisor 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 $$\sum_p\frac1p=+\infty$$
 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

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

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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.