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How do I prove$$\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\Lambda(k)}{k}-\ln(n)=-\gamma $$

Where $\Lambda(k)$ is the Von-Mangoldt function, and gamma is the euler gamma constant

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Theorem 424 in Hardy and Wright, page 348. – Will Jagy Nov 14 '12 at 4:08
Thanks, can I get a link, I don't have a copy of the book – boby Nov 14 '12 at 4:12
That's why there are libraries. – Gerry Myerson Nov 14 '12 at 4:43
I cant get to a library, that is why I asked for a link – boby Nov 14 '12 at 6:14
See too 3.3 of this (.ps) paper. – Raymond Manzoni Nov 15 '12 at 0:28

Xavier Gourdon and Pascal Sebah (in $3.3$ of "Collection of formulae for Euler’s constant $\gamma$") propose to use this formula : $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{k\ge1}\frac {\Lambda(k)}{k^s},\quad s>1$$

that they rewrite as : $$\zeta(s)+\frac{\zeta'(s)}{\zeta(s)}=-\sum_{k\ge1}\frac {\Lambda(k)-1}{k^s},\quad s>1$$ before taking the limits as $s\to 1$ and deducing : $$2\,\gamma=-\sum_{k\ge1}\frac {\Lambda(k)-1}k$$ (the existence of the limit at the right could be questionable...)

The limit at the left may indeed be obtained using (for $|\epsilon|\ll 1$ and $\gamma_1$ a Stieltjes constant) :

  • $\displaystyle \zeta(1+\epsilon)=\frac 1{\epsilon}+\gamma-\gamma_1 \,\epsilon+O(\epsilon^2)\ $ and
  • $\displaystyle \zeta'(1+\epsilon)=-\frac 1{\epsilon^2}-\gamma_1+O(\epsilon)$

so that $\ \displaystyle \frac{\zeta'(1+\epsilon)}{\zeta(1+\epsilon)}=-\frac 1{\epsilon}\frac{1+\gamma_1\,\epsilon^2}{1+\gamma\,\epsilon}+O(\epsilon)=-\frac 1{\epsilon}+\gamma+O(\epsilon)\ $
and $\ \displaystyle \lim_{\epsilon\to 0} \zeta(s)+\frac{\zeta'(s)}{\zeta(s)}=2\gamma\ $ as required.

To get your limit we will just need the additional definition of $\gamma$ : $$\gamma=\lim_{n\to\infty}\left(-\log(n)+\sum_{k=1}^n\frac 1k\right)$$ rewrite the previous limit as : $$-2\,\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac {\Lambda(k)-1}k\right)$$ and combine these two results to conclude : $$-\gamma=\lim_{n\to\infty}\left(-\log(n)+\sum_{k=1}^n\frac {\Lambda(k)}k\right)$$

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If you can show $$\sum_{n\le x}\log n=x\log x+O(x)$$ and $$\sum_{n\le x}\log n=\sum_{n\le x}[x/n]\Lambda(n)$$ then you can get $$\sum_{n\le x}{\Lambda(n)\over n}=\log x+O(1)$$ which isn't as strong as what you want but is in the same ballpark. This is the essence of Hardy & Wright Theorem 424 (6th edition).

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Got your note. Evidently I had a brain relocation, the fifth edition does not mention $\gamma$ either, just what you quote. So now we've got to wonder where the $\gamma$ came from, and whether it's correct. – Will Jagy Nov 15 '12 at 0:09

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