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I'm curious about the Dirichlet series $$F(s) = \sum_{n = 1}^\infty \frac{H_n}{n^s}$$ of the sequence $H_n = \sum_{k = 1}^n \frac{1}{k}$ of harmonic numbers. Its abscissa of convergence is $1$. What properties does it have? Does it have analytic continuation? Does it have a closed-form expression? What is its asymptotic behavior at $s = 1$? (It diverges there.) Does it satisfy a functional equation? I can't find any information about it, nor can I answer any of these questions.

Note that $$F(m) = \frac{m+2}{2}\zeta(m+1)-\frac{1}{2}\sum_{k = 1}^{m-2} \zeta(m-k)\zeta(k+1)$$ for any integer $m>1$. That's about all I know.

Oh, and for obvious reasons, $F(s) - 1$ is asymptotic to $\frac{H_2}{2^s}$ as $s \to \infty$.

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    $\begingroup$ if you write $H_n =\log n + \gamma +c_n$ we know that $c_n=O(1/n)$ so $\sum c_n n^{-s}$ has abscissa at most $0$ while of course, we know $\sum \log n /n^s$ etc; similarly we can extend the abscissa by using more asymptotic terms; the asymptotic behavior at $1$ is obviously given by $-\zeta'+\gamma \zeta$ since the remainder series is analytic there $\endgroup$
    – Conrad
    Commented Nov 1, 2023 at 3:06
  • $\begingroup$ I believe the singularity at $0$ is a second-order pole (breaking the series into regimes $n\sim 2^k$ and using $H_n\sim\log n$, we obtain something like the differentiated geometric series formula), though I don't see how to transform this into an asymptotic expansion // EDIT: this is clearly a consequence of the previous comment as well $\endgroup$
    – Ben Johnsrude
    Commented Nov 1, 2023 at 3:45
  • $\begingroup$ What motivated this question? The harmonic numbers seem to be a curious choice for coefficients of a Dirichlet series. $\endgroup$
    – Anurag Sahay
    Commented Nov 1, 2023 at 11:28
  • $\begingroup$ Well, $H_{\lfloor x \rfloor}$ is the summatory function of $\frac{1}{n}$, so $\zeta(s+1) = s \int_1^\infty\frac{H_{\lfloor x \rfloor}}{x^{s+1}}dx$. I was considering the integral $\int_1^\infty \frac{H_{\lfloor x \rfloor}-\log x -\gamma}{x^s}dx$, which equals $G(s) = \frac{1}{s-1}\left(\zeta(s)-\gamma-\frac{1}{s-1}\right)$, and the sum in question, or @Conrad's version (which I had already considered) is a ``discrete integral'' version of that. Note that $G(1) = -\gamma_1$. $\endgroup$
    – Jesse Elliott
    Commented Nov 1, 2023 at 20:40
  • $\begingroup$ You can do the same with any function $f(x)$ (e.g., $f(x) = g(x)-h(x)$ where $g(x)\sim h(x)$). The integral $\int_1^\infty \frac{f(x)}{x^s}dx$ is something people study, but then, why not also the sum $\sum_{n = 1}^\infty \frac{f(n)}{n^x}$? Take $f(x) = \operatorname{li}(x)-\pi(x)$ and you already get something interesting. $\endgroup$
    – Jesse Elliott
    Commented Nov 1, 2023 at 20:52

1 Answer 1

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Let me put my comment in an answer since it seems to solve the problem at least in the analytic continuation, poles, residues etc way.

Fix a level $k \ge 1$, then using the asymptotics $H_n=\log n+\gamma + \sum_{1 \le j \le k}c_jn^{-2j} +r_k(n)$ with $r_k(n)=O_k(n^{-2k})$, we can write:

$$F(s)=-\zeta'(s)+\gamma \zeta(s)+ \sum_{1 \le j \le k}c_j\zeta(s+2j)+R_k(s)$$ valid for $\Re s >1$ where $R_k$ is analytic on $\Re s >1-2k$. Hence he RHS above is meromorphic on $\Re s >1-2k$ giving the continuation of $F$ there with poles at $1-2j, j \ge 0$

Unfortunately, the $H_n$ representation is asymptotic only (so not valid as a series pointwise) since the coefficients $c_k, r_k(n)$ blow up with $k$, so the same happens with the representation of $F$ in the sense that we cannot write it as a series of analytic functions, just an equality at each level $k$ with remainder $R_k$ which is analytic and about which we can find some bounds depending on $k,s$ of course but whose coefficients as a Dirichlet series can be very large in $k$ although of course bounded by $C_kn^{-2k}$ in $n$

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