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edited Jul 18, 2017 at 12:31

$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$
$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :
so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(s)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$, having a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.
$P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.

Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.

For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.

$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(s)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$, having a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.

Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.

For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.

$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :

$P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.

Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.

For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.

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created Jul 18, 2017 at 11:56

$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

$P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(s)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$, having a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.

Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.

For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.