Prime numbers and limit(?) Can someone help me to prove the following:
$$\lim_{x\to\infty}(\sum_{p\leq x}\frac{1}{p}-\log(\log(x)) -C)=0$$
Where $C$ is a proper constant. 
Thank you...
 A: This sketch is adapted from Apostol's proof that for $x\geq 2$ there is a constant A such that 

$$\sum_{p\leq x}\frac{1}{p} = \log\log x + A + O(\frac{1}{\log x}) $$

Since $\lim_{x\to \infty}\frac{1}{\log x} = 0$ this would give your result...if you will accept an A instead of a C.
The proof is essentially the same as that in Landau's Handbuch at pages 100-102 which is probably the same as Mertens'. 
The proof below is an outline, since the footnotes require separate nontrivial proof and I have left out some intermediate integration steps. Note, $\log^n t =(\log t)^n.$
Proof. Let $A(x) = \sum_{p\leq x} \frac{\log p}{p}$ and let
$a(n) = 1 $ for $n$ prime and $a(n) = 0$ for $n$ nonprime. 
Then $\sum_{p \leq x} \frac{1}{p} = \sum_{n \leq x} \frac{a(n)}{n}$ and 
$A(x) = \sum_{n \leq x} \frac{a(n)}{n}\log n.$
Let $f(t) = \frac{1}{\log t}$ in Abel's identity$^1;$ since $A(t) = 0$ for $t<2,$
$$\sum_{p\leq x}\frac{1}{p}= \frac{A(x)}{\log x} + \int_2^x \frac{A(t)}{t\log^2t}dt. \hspace{20mm}(1)$$ We need the ancillary result$^2$ which gives us that 
$A(x) = \log x + R(x),$ with $R(x) = O(1).$
Applying this to (1) we have 
$\sum_{p\leq x}\frac{1}{p}= \frac{\log x+ O(1)}{\log x}+\int_2^x\frac{\log t + R(t)}{t \log^2 t}$
$ = \sum_{p\leq x}\frac{1}{p} = 1 + O(\frac{1}{\log x}) + \int_2^x\frac{dt}{t\log t} + \int_2^x \frac{R(t)}{t \log^2 t}dt \hspace{20mm}(2)$
The last two integrals on the right:
$\int_2^x \frac{dt}{t\log t}= \log\log x - \log \log 2;$
$\int_2^x \frac{R(t)}{t\log^2 t} =\int_2^\infty \frac{R(t)}{t\log^2 t}dt - \int_x^\infty \frac{R(t)}{t\log^2t}dt$
and $\int_x^\infty\frac{R(t)}{t\log^2 t}dt =  O( \int_x^\infty \frac{dt}{t\log^2 t} )= O(\frac{1}{\log x}).$ 
We can write (2) as 
$$\sum_{p\leq x}\frac{1}{p}= \log\log x +O(\frac{1}{\log x}) + A$$
in which $A = 1 - \log \log 2 + \int_2^\infty \frac{R(t)}{t \log^2 t}dt.$ 

$^1$For f smooth on $[y,x],~~ 0 < y < x,$ we can write
$\sum_{y< n \leq x}a(n)f(n)= A(x)f(x) - A(y)f(y) - \int_{y}^x A(t)f'(t)dt.$ 
$^2~$ $\sum_{p\leq x}\frac{\log p}{p}= \log x + O(1) $ for $x\geq 1.$
