Behavior of inverse of $f(s)=\frac{1}{H_n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\frac{1}{i}$

Question

Take a function $f$ defined as follows with $H_n$ referring to Harmonic number

$$f(s,n)=\frac{1}{H_n}\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\frac{1}{i}$$

Suppose $g(\epsilon,n)$ is the inverse of this function, ie $f(g(\epsilon,n),n)=\epsilon$. I'm interested in behavior of $g(\epsilon,n)$ for fixed small $\epsilon>0$ as $n\to \infty$.

Plotting various values, it seems $g(10^{-6},n)\lesssim 10n$. Can someone see a way to explain this analytically?

Edit following Sal's suggestion, problem above can be rewritten as integral for large $n$, and the inverse in $s$ seems to grow (sub)linearly as well

$$f(s,n)\approx \frac{1}{\log n}\int_{i=1}^n \left(1-\frac{1}{i}\right)^{s} \frac{1}{i}$$

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Edit Trying to match Katsurda approximate expression, it seems to give good results for x<25 after which it diverges rapidly

$$\sum_{j=2}^\infty \frac{(-x)^j}{j!}\zeta(j) \approx x(\log x + 2 \gamma -1)$$

After differentiating expression I get

$$\sum_{j=1}^\infty -\frac{(-x)^j}{j!} \approx \log x+2\gamma$$

Answer 1

First note that $$ \sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^s \frac{1}{i}} = \int_1^n {\left( {1 - \frac{1}{x}} \right)^s \frac{{dx}}{x}} + \mathcal{O}(1)\frac{1}{n}\exp \left( { - s\log \left( {\frac{n}{{n - 1}}} \right)} \right) $$ as $n\to +\infty$. Now \begin{align*} & \int_1^n {\left( {1 - \frac{1}{x}} \right)^s \frac{{dx}}{x}} = \int_{1/n}^1 {\exp \left( { - s\log \left( {\frac{1}{{1 - t}}} \right)} \right)\frac{{dt}}{t}} = \int_{\log \left( {\frac{n}{{n - 1}}} \right)}^{ + \infty } {\frac{{e^{ - sw} }}{w}\frac{w}{{e^w - 1}}dw} \\ & = \int_{\log \left( {\frac{n}{{n - 1}}} \right)}^{ + \infty } {\frac{{e^{ - sw} }}{w}(1 + \mathcal{O}(w))dw} \\ & = E_1 \!\left( {s\log \left( {\frac{n}{{n - 1}}} \right)} \right) + \mathcal{O}(1)\frac{1}{s}\exp \left( { - s\log \left( {\frac{n}{{n - 1}}} \right)} \right) \\ & = E_1 \!\left( {s\log \left( {\frac{n}{{n - 1}}} \right)} \right)\left( {1 + \mathcal{O}(1)\max\! \left( {\frac{1}{{s(\left| {\log (s/n)} \right| + 1)}},\frac{1}{n}} \right)} \right) \end{align*} as $n \to +\infty$. Here $E_1$ is the exponential integral (cf. http://dlmf.nist.gov/6.2.E1). Analogously, $$ \frac{1}{n}\exp \left( { - s\log \left( {\frac{n}{{n - 1}}} \right)} \right) = E_1 \!\left( {s\log \left( {\frac{n}{{n - 1}}} \right)} \right)\mathcal{O}(1)\max \!\left( {\frac{1}{{n(\left| {\log (s/n)} \right| + 1)}},\frac{s}{{n^2 }}} \right) $$ as $n\to +\infty$. Consequently, $$ f(s,n) = \frac{1}{{H_n }}\sum\limits_{i = 1}^n {\left( {1 - \frac{1}{i}} \right)^s \frac{1}{i}} \sim \frac{1}{{H_n }}E_1 \!\left( {s\log \left( {\frac{n}{{n - 1}}} \right)} \right), $$ as $n\to +\infty$ uniformly in $s$, provided $s=o(n^2)$. Thus, $g(\varepsilon,n)$ satisfies $$ E_1 \!\left( g(\varepsilon ,n) \log \left( {\frac{n}{{n - 1}}} \right) \right) \sim \varepsilon H_n \Longleftrightarrow g(\varepsilon ,n) \sim \frac{1}{\log \left( {\frac{n}{{n - 1}}} \right)}E_1^{ - 1} \!\left( {\varepsilon H_n } \right) . $$ Note that $\log \left( {\frac{n}{{n - 1}}} \right) \sim \frac{1}{n}$ for large $n$.

It is possible to show by asymptotic inversion that $$ E_1^{ - 1} (x) = - \log x - \log ( - \log x) - \frac{{\log ( - \log x) - 1}}{{\log x}} + \mathcal{O}\!\left( {\frac{{(\log ( - \log x))^2 }}{{\log ^2 x}}} \right), $$ as $x\to 0+$, and $$ E_1^{ - 1} (x) \sim e^{ - x - \gamma } + e^{ - 2x - 2\gamma } + \frac{5}{4}e^{ - 3x - 3\gamma } + \cdots $$ for moderate or large values of $x>0$.

Answer 2

This answer applies only to the case $s$ is large but fixed and in the limit $n \to \infty.$ (That is, $s$ proportional to $n$ is not allowed.) It may be of limited applicability for your problem. Write $$ (1-1/k)^s = (1-1/k)^{k s/k} \approx e^{-s/k} $$ Then $$ \sum_{k=1}^n(1-1/k)^s\frac{1}{k} \approx \sum_{k=1}^n \frac{e^{-s/k}}{k} = \sum_{k=1}^n \frac{1}{k}(1+\sum_{m=1}^{\infty} \frac{(-s/k)^m}{m!})=$$ $$=H_n + \sum_{m=1}^\infty \frac{(-s)^m}{m!} \sum_{k=1}^{n} k^{-(m+1)} $$ $$ \approx H_n + \sum_{m=1}^\infty \frac{(-s)^m}{m!} \zeta(m+1) $$ where in the last line we let $n \to \infty$ from the penultimate line. However, it can be shown (Katsurada) that $$ \sum_{j=2}^\infty \frac{(-x)^j}{j!} \zeta(j) \approx x(\log{x} + 2 \gamma -1) - \zeta(0) + O(x^{1/4}\exp{(-2\sqrt{\pi}x))} $$ Take a derivative this expression, use the asymptotic formula for the harmonic number $H_n$, and collect results: $$ \frac{1}{\log{n}}\sum_{k=1}^n(1-1/k)^s\frac{1}{k} \approx 1-\frac{\log{s}+ \gamma}{\log{n}} $$ This is easily inverted for whatever $s$ or $n$ you have, with the caveat that $s$ is large.

As an example, for $s=20$ and $n=10000,$ the last equation's left hand side is 0.6096 and the RHS approximation is 0.6121.

References: 'Power Series with the Riemann Zeta-function in the Coefficient,' M. Katsurada, Proc. Japan. Acad. 72 Ser. A (1996) p 61-63

'On Mellin-Barnes Type of Integrals & Sums Associated with the Riemann Zeta-function,' M. Katsurada, Public De L' Institut Mathematique 62 (76) 1997, 13-25.

Answer 3

Potential path to solution:

1. Approximate the sum using integral and $p>1$ $$\sum_{i=1}^n \left(1-\frac{1}{i}\right)^s\frac{1}{i}\approx \sum_{i=1}^n \left(1-\frac{1}{i^p}\right)^s\frac{1}{i^p} \approx \int_{1}^n \left(1-\frac{1}{i^p}\right)^s\frac{1}{i^p} di$$

2. Solve for the integral, in this answer

$$(p-1)\int_1^n \left(1-i^{-p}\right)^s \,i^{-p}\, di=\frac{\Gamma \left(2-\frac{1}{p}\right) \Gamma (s+1)}{\Gamma \left(s-\frac{1}{p}+2\right)}-n^{1-p} \,\, _2F_1\left(\frac{p-1}{p},-s;2-\frac{1}{p};n^{-p}\right)$$

For $p=1.01, n=1000, \epsilon=10^{-2}$, substituting closed form for the integral and inverting numerically gives a close match to $g(\epsilon, n)$:

1786.32 Exact 
1785.42 Integral approximation

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Now to get the answer, one would need to take limit of $p\to 1$ and further simplify to get a closed form for the inverse.