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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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    See too 3.3 of this (.ps) [paper](http://numbers.computation.free.fr/Constants/Gamma/gammaFormulas.ps).2012-11-15

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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.2012-11-15