So the question is, Is there a proof for F(x) = O(x^(1/2-r))?

[eljose79]

let note f(x)=O(g(x)) this f(x)<MG(x) being M a constant then would it be true?..

If f(n)=o(n^u) then Sum(1<n<x)f(n)=O(n^u+1) adn Int(1,x)dnf(n)=O(n^u+1)

Another question let be a(n)n^-s and b(n)n^-s two Dirichlet series so a(n)<b(n) for each n then if b(n)n^-s converges for a number Re(a)>1/2lso the series a(n)n^-s converges for Re(a)>1/2

 

[shmoe]

let note f(x)=O(g(x)) this f(x)<MG(x) being M a constant then would it be true?..

If f(n)=o(n^u) then Sum(1<n<x)f(n)=O(n^u+1) adn Int(1,x)dnf(n)=O(n^u+1)

The n in your bounds for the sum and integral should be an x. Also, when u=-1, you get log(x) for a bound, not a constant.

Another question let be a(n)n^-s and b(n)n^-s two Dirichlet series so a(n)<b(n) for each n then if b(n)n^-s converges for a number Re(a)>1/2lso the series a(n)n^-s converges for Re(a)>1/2

Hi, assuming both sequences are non-negative, then this looks fine.

 

[eljose79]

Another question let be F(x)=Sum(n<x)1/n^rthen does exist an r so:

F(x)=O(x^1/2-r)?..where i could find a proof of that?...thanks.

 

[shmoe]

Another question let be F(x)=Sum(n<x)1/n^rthen does exist an r so:

F(x)=O(x^1/2-r)?..where i could find a proof of that?...thanks.

No there doesn't. If r is not 1 then:

$$\sum\limits_{n<x}\frac{1}{n^r}=O(x^{1-r})$$

So your asking if there is an r where $$1-r\leq 1/2-r$$
Which is of course false.

If r=1, then again, no luck since log(x) is not $$O(x^{-1/2})$$.