Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was just reading this question and answer: How will this equation imply PNT and it raised a whole new question:

Given that $\sum_{n\le x} \Lambda(n)=x+o(x)$, prove that $$\sum_{n\le x} \frac{\Lambda(n)}{n}=\log x-\gamma +o(1),$$ where gamma is the Euler constant.

The above question is for the bounty. The full version above has been bothering me. Gerry Myerson pointed out we can prove $\sum_{n\leq x } \frac{\Lambda(n)}{n} =\log x +O(1)$ from only chebyshev estimate. But this is not my question.

My Attempts: I tried partial summation I end up with something like $\sum_{n} \frac{\psi(n)-n}{n^2}$. for that to converge $o(n)$ isn't strong enough. To do it my way, I would need to assume $$\psi(x)-x = O\left(\frac{x}{(\log x)^{1+\epsilon}}\right)$$ so that it converges. (Otherwise it could be as big as $\log x$ which is no good)

Can we prove the above estimate using only the basic prime number theorem $\psi(x)-x=o(x)$? Why or why not? Thank you!

Please note that my question is about a subtlety with converge, and blunt partial summation doesn't seem to work.

share|improve this question
    
What you have in equ 2 would be $-2\gamma+o(1)$, not $o(1)$. –  anon Oct 28 '11 at 22:09
    
@anon: you are correct. –  Math Student Oct 29 '11 at 13:19

3 Answers 3

A few years ago, I assigned something a little weaker, $$\sum_{n\le x}{\Lambda(n)\over n}=\log x+r(x){\rm\ with\ }|r(x)|\le2$$ and it turned out to be the hardest problem on the assignment. Here's the proof I eventually wrote up.

Let $s=[x]$. Note $$\sum_{n\le x}{\Lambda(n)\over n}=\sum_{n\le s}{\Lambda(n)\over n}$$ and $0\le\log x-\log s\lt1$. Now

$$\log s!=\sum_{m\le s}\log m=\sum_{p^r\le s}\log p\sum_{m\le s,p^r\mid m}1=\sum_{p^r\le s}\log p[s/p^r]=\sum_{n\le s}\Lambda(n)[s/n]$$

$$=\sum_{n\le s}\Lambda(n)\left({s\over n}-\left\lbrace{s\over n}\right\rbrace\right)=s\sum_{n\le s}{\Lambda(n)\over n}-\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace$$

Now $$\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace\le\sum_{n\le s}\Lambda(n)\le Cs$$ for some constant $C$ by the Prime Number Theorem (which is actually overkill, as one can prove without PNT that that last sum is bounded by $2s$ for all $s$). Also, comparing $\log s!=\sum\log n$ to $\int_1^s\log t\,dt$ we get $$\log s!=s\log s-s+b(s){\rm\ with\ }|b(s)|\lt\log s+1$$ (cf. Stirling's formula). Putting it together, $$\sum_{n\le x}{\Lambda(n)\over n}=(1/s)\log s!+(1/s)\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace=\log s-1+C+b(s)/s=\log x+r(x)$$ with $|r(x)|\le2$ since $C\le2$.

EDIT: The stronger result is proved in the textbook by Bateman and Diamond, Analytic Number Theory, pages 100-102. First they prove that, with the usual definitions, $\psi(x)\sim x$ implies $M(x)=o(x)$, where $M(x)$ is the summatory function for the Möbius $\mu$-function. Then from the result on $M(x)$ they deduce the result you want. It's a bit long to write out as I'd have to explain the notation introduced earlier in the chapter.

MORE EDIT: It's also Proposition 3.4.4 of Jameson's textbook, The Prime Number Theorem, and it's proved on page 91 of Tenenbaum and Mendes France, The Prime Numbers and Their Distribution, in the Student Mathematical Library series of the American Math Society. Again, the proofs require too much previously developed material for me to attempt to write them out here.

share|improve this answer
1  
I already know this, this is not my question. What you have shown is that $$\sum_{n\leq x} \frac{\Lambda(n)}{n} =\log x +O(1)$$ where the constant must be less then $2$. This is one of mertens estimates. What I want is significantly stronger, and that is $$\sum_{n\leq x} \frac{\Lambda(n)}{n} =\log x -\gamma +o(1).$$ This last estimate implies the prime number theorem, so you cannot hope to prove it without using something that strong. –  Math Student Oct 29 '11 at 13:22
9  
You have a funny way of saying, "Thank you for being the only person to post an answer to my question, and I apologize for forcing you to read my mind by not being clear in my question about what I already know." –  Gerry Myerson Oct 29 '11 at 21:54
    
I have put all my reputation for a bounty, and edited more. –  Math Student Oct 30 '11 at 21:31
1  
Many questions have many answers - a single answer won't deter others, if they have something to say. And a polite comment, thanking someone for the effort while pointing out what needs to be done, would encourage others to participate. More flies with honey.... –  Gerry Myerson Oct 30 '11 at 21:41
    
Thanks for the answer, which is interesting irrespective of the degree to which it settles the question on PNT. –  zyx Oct 30 '11 at 21:56

If you add $\frac 1n$ to your $\Lambda(n)$, the estimate in the condition does not change because $H(x)\sim\log x = o(x)$.

But the desired conclusion changes by a constant of $\frac{\pi^2}{6}$.

Therefore, your condition is not sufficient to prove the conclusion.

share|improve this answer
2  
$\Lambda(n)$ isn't just any old function satisfying the summation condition, it is specifically the von Mangoldt function, and one is permitted to use facts about it in addition to the summation condition. Did you look at the references I cited in the edits to my answer? –  Gerry Myerson Nov 3 '11 at 12:49
1  
@GerryMyerson I know what $\Lambda$ is, my interpretation of the grey question of the OP is that one is not permitted to use other facts about it and I pointed out that this does not work. My interpretation might well be wrong, but I don't see how the references in your answer would help me to read the mind of the OP. –  Phira Nov 3 '11 at 15:37
    
Nothing will help you to read the mind of OP, and for all I know your interpretation of the question is right, and mine is wrong. The references will help you (in the unlikely event that you need any such help) see how to answer the question that I think OP is asking. –  Gerry Myerson Nov 3 '11 at 21:49
    
@Phira: This is a good point, so I give you +1, but it is not exactly what I am looking for. –  Math Student Nov 5 '11 at 11:32

By definition

$$\gamma=\sum_{n\le x}{1\over n}-\log x+o(1).$$

So we see that $$\log x +\gamma + o(1)=\sum_{n\le x}{1\over n}.$$

So it is sufficient to show that $$\sum_{n\le x}{\Lambda(n)-1\over n}=-2\gamma+o(1).$$

Then using Möbius inversion in the form

$$G(x)=\sum_{n\le x}F\left({x\over n}\right)\iff F(x)=\sum_{n\le x}\mu(n)G\left({x\over n}\right)$$

and the asymptotic estimate

$$\sum_{n\le x}\bigg(\log n-d(n)\bigg)=-2\gamma x+O(\sqrt{x})\quad (*)$$

we get the desired result. Note: you'll end up using the PNT in a $\mu$ function estimate during the Möbius inversion, this is how you use your given, and the given estimate is where you save over just using the PNT by itself. As @Phira notes: this is impossible.

Note: $(*)$ can be proved pretty easily using standard summation techniques, the first term is $\log[x]!=x\log x -x+O(\log x)$ which is just partial summation and the second is $x\log x +(2\gamma-1)x+O(\sqrt{x})$ by using Dirichlet's hyperbola method.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.