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Let $\pi_k(n) $ be the number of numbers with k prime factors (repetitions included) less than or equal to n.

If we take the sums:

$z_1(s) = \sum_{n= 1}^\infty \frac{1}{(p_{1,n})^s},~ z_2(s) = \sum \frac{1}{(p_{2,n})^s},~ z_3(s) = \sum \frac{1}{(p_{3,n})^s},~ . . .,$

in which $\sum \frac{1}{(p_k)^s} $ is the sum of inverses of numbers with k factors, repetitions included, then the sum of these and 1 is:

$$1+ \sum_k \sum_n \frac{1}{(p_{k,n})^s} = \zeta(s) = \sum \frac{1}{n^s}.$$

If $s> 1,~ \zeta(s)$ converges and so each of the $z_k(s)$ converge.

The distribution of the classes $\pi_k(n)$ is Poisson-like for large n--most numbers have relatively few prime factors. And while there is no obvious relationship between $\pi_k(n)$ and $z_k(s)$, it seems that since there is no reason to say (for large numbers) inverse primes are larger or smaller than inverse almost-primes, the size of $z_k$ would be controlled by the cardinality of $\pi_k.$ As s gets closer to 1, we need a larger number of terms in $z_k$ to approximate the actual limit at $\infty$, which means the distribution will (very slowly) begin to resemble the distribution of $\pi_k(n).$

Maybe this is a good illustration of the question for k = 1 (primes):

Might it be true for some sequence $s(n_j) = s_{n_j}> 1$ sufficiently small, with $s_{n_{j+1}} < s_{n_j}$

$$\frac{1}{\zeta(s_{n_j})-1} \sum_{m = 1}^\infty \frac{1}{(p_m)^{s_{n_j}}} \approx \frac{\pi(n_j)}{n_j}$$


A quick calculation of $ z_k$ for k = 1 - 6 and $s = 2,1.5,1.3,1.2$ if plotted show that k = 2+ are growing slowly relative k = 1.

k = $1\hspace{5mm} 2 \hspace{8mm} 3 \hspace{8mm}4 \hspace{9mm} 5 \hspace{9mm} 6$

.4522 .1407 .0385 .0100 .0025 .0007 $\hspace{5mm}$($~\sum \approx .6449$ vs.$ ~\zeta(2)=.6449 + 1$)

.8493 .4470 .1924 .0748 .0278 .0100 $\hspace{5mm}$ ($\sum \approx 1.6$ vs. 1.61+1 actual $\zeta(1.5)$)

1.2012 .8384 .4526 .2137 .0932 .0386 $\hspace{5mm}$ ($\sum \approx 2.83$ vs. 2.93+1)

1.50$\hspace{4mm}$ 1.25$\hspace{3mm}$ .771 $\hspace{3mm}$ .404 $\hspace{1mm}$ .189 $\hspace{1mm}$ .085 $\hspace{10mm}$ ($\sum \approx 4.2$ vs. 4.91+1)

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I'm not sure I fully understand your question, but it seems that it might be relevant that $z_1(1)$ diverges as $\log\log n$ (see e.g. Wikipedia) whereas $\zeta(1)$ diverges as $\log n$. –  joriki Sep 26 '12 at 7:05
See also this question. –  joriki Sep 26 '12 at 7:13
+1 Interesting. Maybe you are interested in this or references therein... –  draks ... Sep 27 '12 at 14:03
@draks: Thanks!--I went back to the original question in MO. Ultimately same sort of question. –  daniel Sep 27 '12 at 14:22
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