What is the asymptotic form for the sum of the reciprocals of the first $n$ primes? It is well known that the sum of the reciprocals of the primes $p$ less than or equal to a maximum value $n$ is asymptotic to $\ln \ln n$:
$$\sum_{p \leq n} \frac{1}{p} \sim \ln \ln n.$$
(The next subleading term on the RHS is the Meissel–Mertens constant.)
But what's the asymptotic form for the sum of the reciprocals of the first $k$ primes? That is, if $p_k = \{2, 3, 5, 7, 11, \dots\}$ is the sequence of primes, then what is the asymptotic form of
$$\sum_{k=1}^n \frac{1}{p_k}?$$
The latter expression seems to me to be a closer variant of the famous asymptotic relation
$$\sum_{k=1}^n \frac{1}{k} \sim \ln n$$
for the harmonic numbers.
 A: Well, $\frac{m \log m}{2} \le p_m \le 2 m \log m$. So $\sum_{k=1}^m \frac{1}{p_k}$ satisfies
$$\ln \ln (m \log m/2) \approx \sum_{p \le \frac{m \log m}{2}} \frac{1}{p} \le \sum_{k=1}^m \frac{1}{p_k} \le
\sum_{p \le 2m \log m} \frac{1}{p} \approx \ln \ln (2m \log m )$$.
However, $\ln \ln (m \log m/2)$ $\approx \ln \ln m \approx$ $\ln \ln (2m \log m )$, yielding 
$$\sum_{k=1}^m \frac{1}{p_k} \approx \ln \ln m.$$
And it makes sense: From analysis (forget about primes and their distributions for a moment)
$\sum_{k=1}^N \frac{1}{k \ln k} \approx \ln \ln N$ (and $p_k \approx k \ln k$); $\sum_{k=1}^N \frac{1}{k \ln k \ln \ln k} \approx \ln \ln \ln N$  while $\sum_{k=1}^{\infty} \frac{1}{k \ln^2 k}$ is finite. In fact, so is $\sum_{k=1}^{\infty} \frac{1}{k (\ln k) (\ln \ln k)^2}$. And even, $\sum_{k=1}^{\infty} \frac{1}{k (\ln k) (\ln \ln k) (\ln \ln \ln k)^2}$ is finite too.
If this is still surprising to you, let $Y(x) = e^{e^x}$, then $dY/dx = Y \ln Y$. And let $Z = e^{e^{e^x}}$. Then $dZ/dx = Z \ln Z \ln \ln Z$. 
