On the relation between the asymptotics of a Dirichlet series' coefficients and the series' analytic continuabilitynte Ee 0dcalde
There is a wonderful series of articles by Flajolet et. al. about Mellin Transforms and the asymptotic analysis of generating functions. In particular, on page 45 of the article Mellin Transforms and Asymptotics: Harmonic Sums, they state the following result:
Proposition 6 (Growth of Special Dirichlet Series): Let $\\lambda_{k}$ and $\\mu_{k}$ admit asymptotic expansions in descending powers of $k$ as $$\\lambda_{k}\\sim\\sum_{r=0}^{\\infty}\\frac{a_{r}}{k^{\\alpha_{r}}}$$ $$\\mu_{k}\\sim k^{w}\\left(1+\\sum_{r=1}^{\\infty}\\frac{b_{r}}{k^{\\beta_{r}}}\\right)$$ Then the Dirichlet series $\\sum_{k}\\lambda_{k}\\mu_{k}^{-s}$ can be continued to a meromorphic function $\\Lambda\\left(s\\right)$ in the whole of the complex plane.
Now, Let $V$ be an arbitrary set of infinitely many positive integers, and let: $$\\zeta_{V}\\left(s\\right)\\overset{\\textrm{def}}{=}\\sum_{v\\in V}\\frac{1}{v^{s}}$$
Enumerating the elements of $V$ in increasing order as $v_{1},v_{2},\\ldots$, recall that one way of defining the natural density of $V$ (denoted $d\\left(V\\right))$ is: $$d\\left(V\\right)=\\lim_{n\\rightarrow\\infty}\\frac{n}{v_{n}}$$ In the case where $d\\left(V\\right)$ exists and is positive, this gives the asymptotic $d\\left(V\\right)v_{n}\\sim n$. As such, for: $$\\frac{\\zeta_{V}\\left(s\\right)}{\\left(d\\left(V\\right)\\right)^{s}}=\\sum_{n=1}^{\\infty}\\frac{1}{\\left(d\\left(V\\right)v_{n}\\right)^{s}}$$ I have that $$\\lambda_{k}=1$$ and $$\\mu_{k}=d\\left(V\\right)v_{k}\\sim k$$ which is obtained by taking:$$\\mu_{k}\\sim k^{w}\\left(1+\\sum_{r=1}^{\\infty}\\frac{b_{r}}{k^{\\beta_{r}}}\\right)$$ setting $w=1$, and letting all the $b_{r}$s be $0$. Hence, unless I am mistaken, Proposition 6 implies that $\\zeta_{V}\\left(s\\right)$ extends to a meromorphic function on $\\mathbb{C}$ whenever $V$ has positive natural density.
However, consider the following. Let $\\mathbb{P}$ denote the set of prime numbers. Then, $\\zeta_{\\mathbb{P}}\\left(s\\right)$ is the so-called Prime Zeta Function, which is known to have a natural boundary on the imaginary axis. On the other hand, since $d\\left(\\mathbb{P}\\right)=0$, it follows that $\\mathbb{N}/\\mathbb{P}$ has a well-defined natural density of $1$, and thus, by Proposition 6, that $\\zeta_{\\mathbb{N}/\\mathbb{P}}\\left(s\\right)$ is meromorphic on $\\mathbb{C}$. However, since I can write:$$\\zeta_{\\mathbb{P}}\\left(s\\right)=\\zeta\\left(s\\right)-\\zeta_{\\mathbb{N}\\backslash\\mathbb{P}}\\left(s\\right)$$ it follows that $\\zeta_{\\mathbb{P}}\\left(s\\right)$ is the difference of two meromorphic functions, which forces $\\zeta_{\\mathbb{P}}\\left(s\\right)$ to be meromorphic on $\\mathbb{C}$, which is obviously not correct.
So, where's the error, and how (if at all) can it be rectified? In particular, when, if ever, does the existence of $d\\left(V\\right)$ imply that $\\zeta_{V}\\left(s\\right)$ is meromorphic on $\\mathbb{C}$?
2 Answers
Proposition 6 does not imply that if $V$ has positive natural density then $\\zeta_V(s)$ extends to a meromorphic function. This is because the density assumption is much weaker than the assumptions in Proposition 6. Indeed, if the elements of $V$ are $v_1<v_2<\\dotsb$, then the density assumption says that $v_k$ is asymptotically a constant times $k$, while Proposition 6 requires a full asymptotic expansion of $v_k$ in descending powers of $k$. The latter means that, for any nonnegative integer $R$, there are exponents $w_1>w_2>\\dotsb>w_R$ and coefficients $c_1,c_2,\\dotsc,c_R\\in\\mathbb{R}$ such that $$v_k=c_1 k^{w_1}+c_2 k^{w_2}+\\dotsb+c_{R-1}k^{w_{R-1}}+(c_R+o(1))k^{w_R}.$$ The density assumption implies that we have such a relation for $R=1$, namely $w_1=1$ and $c_1=1/d(V)$ work, but it does not imply the relation for $R=2$. And in fact, for the sequence of non-primes $\\mathbb{N}\\setminus\\mathbb{P}$, the above approximation only holds for $R=1$.
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$\\begingroup$ can you check if I didn't miss some point $\\endgroup$ – reuns 7 hours ago
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$\\begingroup$ @reuns: I trust what you write. Note also that the validity of Prop. 6 is not an issue, the quoted paper sketches the proof. The OP seems to have mixed up the terms "is asymptotic to" and "has asymptotic expansion", and this caused the confusion. $\\endgroup$ – GH from MO 7 hours ago
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$\\begingroup$ @GH: See, this is why I dislike people who sketch proofs and speak sketchily of things. ;) $\\endgroup$ – MCS 7 hours ago
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$\\begingroup$ Done. You're welcome. I already am very well aware of the content you explained. It's incredibly frustrating to get tripped up simply because of other people's careless failures to actually say what they mean, rather than mere mean what they say. $\\endgroup$ – MCS 7 hours ago
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$\\begingroup$ @MCS: I think you are too hard on the authors. They do say "asymptotic expansion", and the definition of this notion is standard: en.wikipedia.org/wiki/Asymptotic_expansion#Formal_definition $\\endgroup$ – GH from MO 6 hours ago
$$F(s)=\\sum_{k}\\lambda_{k}\\mu_{k}^{-s}$$ extends meromorphically with finitely many poles in each strip because $$\\Gamma(s)F(s)= \\int_0^\\infty f(t)t^{s-1}dt, \\qquad f(t)=\\sum_{k}\\lambda_{k} e^{-t \\mu_{k}}$$ where $f$ has an expansion in powers of $t$ near $0$ which makes clear the exact conditions you need for it to hold.
- The expansion $\\mu_k= k^{w}\\left(1+\\sum_{r\\le R}b_{r}k^{-\\beta_r} + o(k^{-\\beta_R})\\right)$ leads to (expanding the $\\exp$) $$e^{-\\mu_k t}= e^{-k^wt} \\exp(- \\sum_{r\\le R} b_r k^{(w-\\beta_r) t}) (1+o(k^{-\\beta_R})) \\\\= e^{-k^wt} \\sum_{j\\le J} c_j t^{d_j} k^{e_j}+ o(t^{d_J}k^{-e_j} e^{-k^wt})$$
- together with the expansion $\\lambda_k = \\sum_{r \\le R} a_r k^{-\\alpha_r} + o(k^{-\\alpha_R})$ it leads to $$f(t) = \\sum_{l\\le L} C_l t^{D_l} \\sum_k k^{E_l} e^{-k^wt}+ o(t^{d_L} \\sum_k k^{e_L} e^{-k^wt})$$
For every $r$, for $L$ large enough we have $t^{d_L} \\sum_k k^{e_L} e^{-k^wt} = o(t^r)$ so the error term won't be a problem.
And from our knowledge of $\\Gamma(s) \\zeta(Bs+C)$, of Mellin inversion and the residue and Tauberian theorem we know that $\\sum_k k^{E_l} e^{-k^wt}$ has an expansion in powers of $t$.
Thus so does $f$ $$f(t) = \\sum_{m \\le M} u_m t^{v_m} + o(t^{v_M}), \\qquad v_m \\to \\infty$$ from which we have our meromorphic continuation to $\\Re(s) > -v_M$ $$\\Gamma(s)F(s)= \\int_0^\\infty (f(t)-\\sum_{m \\le M} u_m t^{v_m} 1_{t < 1})t^{s-1}dt+ \\sum_{m \\le M}\\frac{u_m }{s+v_m}$$
