Proof of $\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1$ I am looking for a proof of $$\lim_{s\to 1^+}\frac{\sum_p p^{-s}}{\ln(s-1)}=-1.$$
If the proof is too long, a direct reference is fine. Here the sum $\sum_p$ denotes the sum over all prime numbers.
EDIT: Perhaps this is one of the examples Mathematica does not help. I put in $s=1.0001$ and calculate 
Sum[(Prime[n])^(-1.0001),{n,1,10000}]/Log(0.0001)
I get $-0.294043$. I even use smaller $s$, and larger upper bound for $n$, but the answer is far from $-1$. If you can get it to close to $-1$, please let me know how you do it.
 A: Since
$$
\lim\limits_{s\to 1^+}(s-1)\zeta(s)=1\tag{1}
$$
it is enough to show that
$$
\lim\limits_{s\to 1^+}\frac{\sum\limits_{p\in P} p^{-s}}{\log\zeta(s)}=1\tag{2}
$$
Using Euler poduct formula we get
$$
\log\zeta(s)
=\log\prod\limits_{p\in P}\frac{1}{1-p^{-s}}
=\sum\limits_{p\in P}\log(1-p^{-s})
=\sum\limits_{p\in P}\sum\limits_{n=1}^\infty\frac{p^{-sn}}{n}\\
=\sum\limits_{p\in P}p^{-s}+\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{p^{-sn}}{n}\tag{3}
$$
We have the following bound
$$
\begin{align}
\left|\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{p^{-sn}}{n}\right|
&\leq\sum\limits_{p\in P}\sum\limits_{n=2}^\infty\frac{1}{n}\left|\frac{1}{p^{sn}}\right|\\
&=\sum\limits_{n=2}^\infty\sum\limits_{p\in P}\frac{1}{n}\frac{1}{|p^{sn}|}\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{\Re(s)n}|p^{\Im(s)n}|}\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{\Re(s)n}}\\
&\leq\sum\limits_{n=2}^\infty\frac{1}{n}\sum\limits_{p\in P}\frac{1}{p^{n}}\\
&\leq\sum\limits_{n=2}^\infty\frac{1}{n}\int_1^\infty\frac{1}{t^n}dt\\
&=\sum\limits_{n=2}^\infty\frac{1}{n}\frac{1}{n-1}\\
&=\sum\limits_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)\\
&=1\\
\end{align}\tag{4}
$$
From $(1)$ we know that $\lim\limits_{s\to 1^+}\log\zeta(s)=\infty$, then from bound $(4)$ and equality $(3)$ it follows that $(2)$ holds. And now we are done!
