The order of magnitude of $\prod_{p \leq N}(1 + p^{-r})$ as $N \to \infty$ Let 
\[
f(N, r) \equiv \prod_{p \leq N} (1 + p^{-r}) \quad 0 < r < 1
\]
where $p$ runs through all the primes less than or equal to $N$.
What is the order of magnitude of $f(N, r)$ as $N \to \infty$?
I think this sounds a very easy problem. But I am doubtful of
what I did.
The obvious plan is to exponentiate, take the logarithm, and then
use the prime number theorem.
So I get
\[
f(N, r) = \exp[\log f(N, r) ] = \exp[\sum_{p \leq N}p^{-r} + O(\sum_{p \leq N} p^{-2r})  ] \asymp \exp[\sum_{p \leq N }p^{-r}].
\]
With the prime number theorem and Abel summation formula, I get
\[
\sum_{p \leq N}p^{-r} \asymp \pi(N)N^{-r} + r \int_{2}^{N}\pi(u)u^{-r - 1} du
\gg N^{1 - r} (\log N)^{-1}.
\]
Thus, I get
\[
f(N, r) \gg \exp[N^{1 - r} (\log N)^{-1}] \quad ?
\]
But my instinct (which is actually not good and often careless) tells me that "$f(N)$ increases exponentially" seems very wrong. 
I used Mathematica and got
\[
\lfloor f(p_{200}, 2/3) \rfloor = \lfloor f(1223, 2/3) \rfloor = 486,
\]
which is not of exponential order at all.
Please direct me to the right path.

Arguing non-rigorously,
$\begin{array}\\
\sum_{p \leq N}p^{-r}
&\approx \sum_{k=1}^{N/\ln(N)} (k \ln k)^{-r}\\
&\approx \int_2^{N/\ln(N)}\dfrac{dx}{(x\ln(x))^r}\\
&\sim \dfrac1{\ln^r(N/\ln(N))}\int_2^{N/\ln(N)}\dfrac{dx}{x^r}
\qquad\text{since }\ln^r(x) \text{ doesn't vary much}\\
&\sim \dfrac1{\ln^r(N/\ln(N))}\dfrac{x^{1-r}}{1-r}|_2^{N/\ln(N)}\\
&\sim \dfrac1{\ln^r(N/\ln(N))}\dfrac{(N/\ln(N))^{1-r}}{1-r}\\
&\sim \dfrac{N^{1-r}}{(1-r)(\ln(N))^{1-r}(\ln(N)-\ln\ln(N))^r}\\
&\sim \dfrac{N^{1-r}}{(1-r)\ln(N)}\\
\end{array}
$
which is about what you got.
According to Wolfy,
for $N = 200\ln(200), r=2/3$,
this is about $4.39$
and
exp of this is
about $80.70$
which is within a factor
of 6 of your computation.
Since the approximations
are moderately crude
with lots of constants
being thrown away,
I don't think that
this is too bad.
