Is there a "closed" formula for the sum $\sum_{n}^{\infty }\frac{\Omega (n)}{n^s}$ in terms of function $\zeta(s)$ (Riemann zeta ) and its derivatives? Here $\Omega (n)$ denote the total number of prime divisors of $n$.
Formula for the sum $\sum_{n}^{\infty }\frac{\Omega (n)}{n^s}$ in terms of the Riemann zeta function
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3$\begingroup$ nearly certainly not, since $\Omega$ is additive, not multiplicative. for some other results, see this. $\endgroup$– mathworker21Commented May 3 at 11:28
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5$\begingroup$ To elaborate on mathworker21's comment: since $\Omega = \mathbf{1} * \mathbf{1}_{PP}$ where $\mathbf{1}$ is the constant function $1$ and $\mathbf{1}_{PP}$ is the indicator of prime powers, it follows that $\sum_{n} \Omega(n)/n^s = \sum_{n} 1/n^s \sum_{p ,\, i \ge 1} 1/(p^{is}) = \zeta(s) F(s)$ where $F(s) := \sum_{p, i\ge 1} 1/p^{is}$. In fact, $F(s)$ can be expressed in terms of $(\zeta(sn))_{n}$, and this goes back to Riemann: $F(s) = \sum_{i \ge 1} F_P(is)$ where $F_P(s) := \sum_{i \ge 1}1/p^s$, and by Möbius inversion, $F_P(s) = \sum_{d\ge 1} \frac{\mu(d)}{d}\log \zeta(sd)$. $\endgroup$– Ofir GorodetskyCommented May 3 at 11:47
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3$\begingroup$ (cont. ) One may express $F(s)$ as $\sum_{d \ge 1} \frac{\phi(d)}{d} \log \zeta(sd)$ with little effort (basically because $\sum_{e \mid d} \mu(e)/e = \phi(d)/d$). See also the wiki page for the Prime zeta function: en.wikipedia.org/wiki/Prime_zeta_function $\endgroup$– Ofir GorodetskyCommented May 3 at 12:54
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2$\begingroup$ For $\Re s\ge 1+\varepsilon$ we have $|\log \zeta(s)|\ll_{\varepsilon} 1/2^{\Re s}$ and $|G(s)| \ll_{\varepsilon} 1/2^{\Re s}$ which implies all your series converge absolutely and uniformly in $\Re s\ge 1+\varepsilon$ so these operations are easily justified. For $\Re s\le 1$ things are of course more subtle. $\endgroup$– Ofir GorodetskyCommented May 3 at 18:16
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2$\begingroup$ @OfirGorodetsky's first expression is equivalently $\sum_{n = 1}^\infty \frac{\Omega(n)}{n^s} = \zeta(s) \sum_{n = 1}^\infty P(ns)$, where $P(s)$ is the prime zeta function. This follows formally from the fact that an arithmetic function $f$ is additive if and only if $\mu * f$ is supported on prime powers $>1$. $\endgroup$– Jesse ElliottCommented May 8 at 7:01
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