Find the smallest $N$ such that $\sum_{k=1}^N\frac{1}{p_k}>\pi$. (The $p_k$'s are the prime numbers.) How to solve the following problem?

Let $\{p_k\}_{k=1}^\infty$ be the set of primes (in increasing order). What is the smallest integer $N$ such that
  $$\sum_{k=1}^N \frac{1}{p_k}>\pi?$$

We know that $\sum_{k=1}^\infty\frac{1}{p_k}$ diverges, so $N$ must exists. Moreover, the sum diverges like $\log\log N$ (very, very slowly) so we could get an $N$ such that $\sum_{k=1}^\infty\frac{1}{p_k}$ will be pretty close to $\pi$.
I tried to solve this problem numerically, but it takes such a long time to compute that I wasn't able to get the result. Is there a clever way to find it?
 A: You can verify that
$$
\begin{align}
\sum_{k=1}^{3\ 260\ 805}\frac{1}{p_k} &= 3.1415926\color{blue}{396}... \\
\pi &= 3.1415926\color{blue}{535}... \\
\sum_{k=1}^{3\ 260\ 806}\frac{1}{p_k} &= 3.1415926\color{blue}{579}...
\end{align}
$$
so
$$N=3\ 260\ 806.$$
A: According to Wikipedia, one has $\sum \limits _{p \leq N} \frac 1 p \geq \log \log (N+1) - \log \frac {\pi^2} 6$. Now, this sum can be rewritten as $\sum \limits _{n=1} ^{\pi(N)} \frac 1 {p_n}$ and since $\pi(N) \leq N$ we have that $\sum \limits _{n=1} ^{N} \frac 1 {p_n} \geq \log \log (\pi(N) + 1) - \log \frac {\pi^2} 6$. In turn, we know that $\pi(N) \geq \frac N {\log N}$, so you get $\sum \limits _{n=1} ^{N} \frac 1 {p_n} \geq \log \log ( \frac N {\log N} +1) - \log \frac {\pi^2} 6$.
The only thing to do now is to impose some inequalities: if we call $C= \pi + \log \frac {\pi^2} 6$ and take into account that $\frac N {\log N} \geq \frac N {\sqrt N}$, we have to look for $N$ such that $\log \log \sqrt N \geq C$, which gives $N \geq \mathbb{e}^{2 \mathbb{e}^C}$. Now, starting from this value (at which the inequality is guaranteed to hold) go backwards bisecting your intervals until you find the minimum such $N$. Since bisecting takes $O(\log N)$ steps, this will be really fast even for an old computer.
