Does $$\sum_{p\in\mathbb P}\frac {( - 1)^{[\sqrt p\,]}}{p}$$ converges ?
I know that the following $\sum_{p\in\mathbb P} \frac{1}{p}$ diverges, we can find proofs on Wikepedia Divergence of the sum of the reciprocals of the primes
Unfortunately this one is more complicated.
I think one needs to use some non-trivial facts about their distribution.
On
Try use Abel summation (http://en.wikipedia.org/wiki/Abel's_summation_formula) with $f(x)=\frac1x$ and $a_n = 0$ if $n\neq p$ and $a_n = (-1)^{[p]}$
With partial summation, or summation by parts, in Riemann-Stieltjes integration by parts (they're all the same, just a choice of name), we get something which makes the distribution question more clear:
$$ \sum_{p \leq x} \frac{(-1)^{\lfloor \sqrt p \rfloor}}{p} = \left(\sum_{p \leq x} (-1)^{\lfloor \sqrt p \rfloor} \right)\frac{1}{x} + \int_1^x \frac{\sum_{p \leq t} (-1)^{\lfloor \sqrt p \rfloor} }{t^2}\mathrm dt,$$
and the first term on the right clearly goes to $0$ as $x \to \infty$. The second term is more exciting, and I do not have a complete answer for it.
If we naively and grossly overestimate the integral, using that there are $t/\log t$ primes up to $t$, then the integral becomes
$$ \int_1^x \frac{1}{t\log t} \mathrm{d}t,$$
which just barely diverges as $x \to \infty$. Really, essentially any savings at all on the naive estimate would give convergence.
If we go to the other side and assume that the parity of $\lfloor \sqrt p \rfloor$ is a random, independent event (which I actually would suspect is true asymptotically), so that it's $\pm 1$ with coin flip probability, then you expect square root cancellation. That is, we might heuristically expect that
$$ \left\lvert\sum_{p \leq x} (-1)^{\lfloor \sqrt p \rfloor} \right\rvert \lessapprox \sqrt{x}.$$
If this is the case, then the integral is bounded above in absolute value by
$$\int_1^x \frac{1}{t^{1.5}}\mathrm{d}t,$$
which trivially converges, and thus leads us to believe that the overall sum converges.
A very reasonable question is, what do we actually know about the distribution of primes in this case? And unfortunately, the answer is that we know extremely little about the distribution of primes along almost any curve that's not linear. We do not know of any nonlinear polynomial that hits infinitely many primes. Relatedly are Landau's 4th problem (is $n^2 + 1$ prime infinitely often?) and Schinzel's Hypothesis H (similar problem about multiple polynomials at once). To be fair, what we are asking for feels weaker than either of those results, but it's still very nontrivial.
So heuristically, it should converge.
What follows are heuristics and numerical evidence supporting the above heuristic for convergence (read: not conclusive, but convincing).
If $2n \leq \sqrt p < 2n+1$, then $\lfloor \sqrt p \rfloor$ is even. Similarly, if $2n + 1 \leq \sqrt p < 2(n+1)$, then $\lfloor \sqrt p \rfloor$ is odd. So we wish we understood numbers of primes in the segments of the form
$$\begin{align} (2n)^2 &\leq p < (2n+1)^2 \\ (2n+1)^2 &\leq p < (2n+2)^2 \end{align}$$
Asymptotically in $n$, we expect there to be $$\pi( (2n+1)^2) - \pi( (2n)^2) \approx \frac{(2n+1)^2}{2\log(2n + 1)} - \frac{(2n)^2}{2\log 2n} \approx \frac{4n}{2\log 2n}$$ primes in the first range, and $$\pi( (2n+2)^2 - \pi ((2n+1)^2 ) \approx \frac{(2n+2)^2}{2\log (2n + 2)} - \frac{(2n+1)^2}{2\log 2n+1} \approx \frac{4n}{2\log 2n}$$ in the second range, and where these approximations become better and better as $n$ increases. Thus we expect the number of positive terms of $(-1)^{\lfloor \sqrt p \rfloor}$ one interval to cancel the negative terms of $(-1)^{\lfloor \sqrt p \rfloor}$ in the subsequent interval, and we should expect more than square root cancellation as $n \to \infty$ (We should expect an extremely high amount of cancellation), as long as we're looking at an enpoint of one of these intervals.
If we suppose that this cancellation really does happen, this means that we might heuristically say that in the sum $\displaystyle \sum_{p \leq m} (-1)^{\lfloor \sqrt p \rfloor}$, the terms below the largest square less than $m$ roughly cancel each other, leaving only those terms after that square to contribute to the sum. How many terms is this? The largest square below $m$ is $\lfloor \sqrt m \rfloor ^2$, so there are $m - \lfloor \sqrt m \rfloor ^2$ terms. To be safe and handle parity concerns, let's say we have twice as many terms, $2(m - \lfloor \sqrt m \rfloor ^2)$.
To get a handle on this size, recall that $\sqrt m = \lfloor \sqrt m \rfloor+ \{ \sqrt m \}$, to that $$2(m - \lfloor \sqrt m \rfloor^2) = 2(m - (\sqrt m - \{ \sqrt m \})^2) = 4\sqrt m \{ \sqrt m \} - 2 \{\sqrt m\}^2 < 4\sqrt m, $$ leading us back to square root cancellation (and this is a pretty generous estimation).
So I feel pretty content in the claim that we should expect $$\left\lvert\sum_{p \leq m} (-1)^{\lfloor \sqrt p \rfloor} \right\rvert \lessapprox 4\sqrt{m}$$ for all $m$ sufficiently large, and that it is almost always much smaller. In fact, I might even believe that this argument could be made rigorous if you wanted to pay attention to enough error terms and carry all the $O(\cdot)$ terms around (which I certainly don't).
For verification, I ran some tests against the first several million primes, and I've included the start of that data here. The
actual valueis the value of the sum, and thenaive valueis assuming square root cancellation. So it seems pretty convincing that we get more than square root cancellation almost always, as expected.