Remainder of a prime series We can see the remainder
$$ R(x) := \sum_{\substack{\nu \geqslant 2 \\ p^{\nu} > x}} \frac{1}{p^{\nu}} $$
goes to $0$ when $x$ tends to infinity but how can we get the " order of magnitude " please ?
I tried the following reasoning : I think I proved for $y, \alpha > 1$, we have
$$ \sum_{p > y} \frac{1}{p^\alpha} \asymp \frac{1}{y^{\alpha-1}\,\ln\,y}, $$
for example by summation by parts and so we get here
$$ R(x) = \sum_{\nu \geqslant 2} \, \sum_{p > x^{1/\nu}} \frac{1}{p^{\nu}} \asymp \sum_{\nu \geqslant 2} \, \frac{1}{(x^{1/\nu})^{\nu-1}\,\ln\left(x^{1/\nu}\right)} = \frac{1}{x\,\ln\,x} \, \sum_{\nu \geqslant 2} \, \nu\,x^{1/\nu},$$
which doesn't work, since the series obtained diverges, so there is a huge mistake, but I can't see where ... Does it go for the first formula (we may have just $\ll$ and not $\asymp$), please?
Thank you in advance.
 A: The first formula is correct, but only in the limit $x \rightarrow \infty$, and certainly not for $x \sim 1$. Trouble is, for large $\nu$, $x^{1/\nu}$ is close to $1$. 
On the other hand, let $q(n)$ denote the number if all $p^{\nu}$, $p$ prime, $\nu \leq 4$, with $p^{\nu} \leq n$. 
Clearly $q(n) \leq |\{(t,k),\, t\geq 4, k \geq 2,\,t\ln{k} \leq \ln{n}\}|$, and there are $\leq \ln_2(n)$ possibilities for $k$ and $\leq n^{1/4}$ for $n$. 
So $q(n) \leq n^{1/4}\ln{n}$. 
So, if $q_1(n)$ is the cardinality of the set of prime powers $\leq n$ that are not prime squares, $q_1(n) \sim 3\frac{n^{1/3}}{\ln{n}}$. 
So if $m_n$ is the $n$-th element that is a prime power not a prime square, then $m_n \sim n^3\ln{n}^3$. Thus 
$$\sum_{m_p > n}{\frac{1}{m_p}} \sim \sum_{p > q_1(n)}{\frac{1}{p^3\,ln^3{p}}} \sim C\frac{q_1(n)^{-2}}{\ln^3{q_1(n)}} \sim \frac{C}{n^{2/3}\ln{n}}.$$
As you showed, the sum of reciprocals of prime squares greater than $n$ is $\sim C\frac{1}{\sqrt{n}}{\ln{n}}$. 
So finally $R(n) \sim \frac{C}{\sqrt{n}\ln{n}}$. 
A: For those who would be interested, I have made an advance and I don't know if we can do better :
$$R(x) = \sum_{p > x} \sum_{\alpha = 2}^\infty \frac{1}{p^\alpha} + \sum_{p \leqslant x} \sum_{\alpha = \lfloor \log x / \log p \rfloor + 1}^\infty \frac{1}{p^\alpha} = \sum_{p > x}\frac{1}{p(p-1)} + \sum_{p \leqslant x} \frac{p^{-\lfloor \log x / \log p \rfloor}}{p-1}  , $$
and so we have the bounds
$$ \frac{1}{x\,\ln\,x} \ll R(x) \ll \frac{1}{\ln\,x}. $$
