Convergence of series $\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n}$ How to prove that $\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n}=2\gamma$?
The question is equivalent to this one:
Prove $\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\Lambda(k)}{k}-\ln(n)=-\gamma $
But that question is still unanswered. Raymond Manzoni does suggest a strategy for that question, but it involves the questionable limit statement $\lim_{s\downarrow1}\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n^s}=\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n}$.
By Hardy and Wright, the partial sums of $\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n}$ are $O(1)$. Also, $\lim_{s\downarrow1}\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n^s}=2\gamma$.
But these two results are not enough to prove that the series is convergent for $s=1$: consider, for example, the Dirichlet series $L(s):=\sum_{n=1}^{\infty}\frac{n^{-i}}{n^s}$. We have $\lim_{s\downarrow1}L(s)=\zeta(1+i)$ and the partial sums of $L(1)$ are $O(1)$, but $L(1)$ does not converge.
 A: Let $g(n):=\left(\sum_{i=1}^n\Lambda(i)\right)-n$. Now, by the Prime Number Theorem, there are constants $c,d>0$ such that $g(n)=\psi(n)-n<d\cdot ne^{-c\sqrt{\log(n)}}$ for all $n>0$. Therefore, $\sum_{i=1}^n\frac{1-\Lambda(i)}{i}=\sum_{i=1}^n\frac{g(i-1)-g(i)}{i}=\sum_{i=1}^n(\frac{g(i)}{i+1}-\frac{g(i)}{i})+\frac{g(1)}{1}-\frac{g(n)}{n+1}$, so since $\frac{g(n)}{n+1}\rightarrow0$ as $n\rightarrow\infty$, our sum converges if and only if $\sum_{i=1}^{\infty}\frac{g(i)}{i(i+1)}$ converges. It suffices to show that it converges absolutely. We have that $$\sum_{i=1}^{\infty}\left|\frac{g(i)}{i(i+1)}\right|<\sum_{i=1}^{\infty}\frac{d\cdot ie^{-c\sqrt i}}{i^2}=d\cdot \sum_{i=1}^{\infty}\frac{1}{ie^{c\sqrt{i}}}<d+d\cdot \int_{i=1}^{\infty}\frac{1}{ie^{c\sqrt i}}\\
=d+d\cdot [-\frac{2}{c^2}e^{-c\sqrt{\log(x)}}(c\sqrt{\log(x)}+1)]_1^{\infty}=d+\frac{2d}{c^2},$$
which is finite. This proves that $\sum_{i=1}^{\infty}\frac{1-\Lambda(i)}{i}$ converges.
Now, take the Dirichlet series $L(s)=\sum_{n=1}^{\infty}\frac{1-\Lambda(n)}{n^s}$. We have proved that $L(1)$ converges, hence $L(s)$ converges for all $s$ with $\Re(s)\geq1$. It also follows that $L$ is continuous on $[1,\infty)$, so $L(1)=\lim_{s\downarrow1}L(s)$. We already know that this is equal to $2\gamma$, so $\sum_{i=1}^{\infty}\frac{1-\Lambda(n)}{n}=2\gamma$.
