Prove $\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\Lambda(k)}{k}-\ln(n)=-\gamma $ 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
 A: 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 :
$$-\gamma=\lim_{n\to\infty}\left(-\log(n)+\sum_{k=1}^n\frac {\Lambda(k)}k\right)$$
A: 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). 
