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

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

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

notebook

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

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

• Not sure about the inverse, but by writing the sum as an integral, I find $f \sim n^{-1}(\ln (n)-H_s)$ for $n \to \infty$, where $H_s$ is a Harmonic number.
– Sal
Jun 23, 2021 at 22:33
• @Sal thanks for the tip. If I approximate $H_s$ with $\log(s)$, this inverts easily to get $g(\epsilon, n)=n e^{-\epsilon n}$, but this function is eventually decreasing with $n$....I wonder if I didn't plot far enough Jun 24, 2021 at 0:25
• The relation $H_s \sim \ln(s) + \gamma$ is for $s \to \infty$. Mathematica says $H_s \sim \pi^2s/6$ for $s \to 0$, from which the inverse (using the approximation via integral) is $s(f)=6(\ln(n)-fn)/\pi^2$
– Sal
Jun 26, 2021 at 0:59
• After playing around with it for a bit, I think a problem here is that $f \to 1$ as $n \to \infty$, independently of $s$ as long as $0<s<1$. I used a slightly different integral than the one in your post: experimentally and without proof, I conjecture both have an error that is not algebraic in $n$ and at least of order $1/\ln n$
– Sal
Jun 27, 2021 at 2:24
• Perhaps a redundant comment, but some people seem not to realize that it is often not necessary to actually compute the inverse of a function, certainly not if you are only interested in plotting it. If the function is given by $y=f(x)$ then simply consider $x=f(y)$ and plot it by swapping $x$ and $y$. Jun 28, 2021 at 15:01

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 $$x>0$$.

• If you plot $$n\left( - \log (\varepsilon H_n ) - \log ( - \log (\varepsilon H_n )) - \frac{{\log ( - \log (\varepsilon H_n )) - 1}}{{\log (\varepsilon H_n )}}\right)$$ with $1\leq n \le 1000$ and $\varepsilon=10^{-6}$ it maches with your graph for $g$ quite well.
– Gary
Jul 8, 2021 at 6:23
• @YaroslavBulatov I revised my answer. It shows an asymptotics for large $n$ that is valid whenever $s$ grows slower than $n^2$.
– Gary
Jul 8, 2021 at 8:06
• Thanks, I've tried with some simulations and it is indeed a good fit. If \epsilon is fixed but $n$ grows, eventually I'll need the second approximation, right? I'm wondering how I would get more terms for the large x version -- wolframcloud.com/obj/yaroslavvb/newton/… Jul 8, 2021 at 22:35
• @YaroslavBulatov I expressed $-\gamma -E_1(z)$ via dlmf.nist.gov/6.6.E2, took the exponential of both sides, expanded the exponential of the power series and finally employed series reversion to solve for $z$.
– Gary
Jul 9, 2021 at 5:09

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.

• aha, so $s\propto n^{1-\epsilon}$....that explains why it looks almost linear with $n$ Jul 1, 2021 at 22:45
• I think there was a small typo in the formula due to Katsurda...can you help me locate the source of Katsurda formula to double-check? Jul 5, 2021 at 22:12
• I've updated the post and notebook with numerical checks of what I think Katsurda is, seems to give close match for x<20, after which it diverges rapidly Jul 5, 2021 at 23:01
• @YaroslavBulatov It's good you are diligent, put it puts me in a quandary. I have papers scattered over three physical locations, and have only been able to check parts of two. I'm disappointed that I cannot put my finger on the paper, or I'd give you the reference. I had only 15 minutes to type in the answer above before the bounty expired, and I used what I had scribbled on a scrap of paper many years ago, and the only designation I wrote was 'Katsurda.' Please allow a couple of days, as I am busy at work (math is my hobby.) Jul 6, 2021 at 16:00
• @YaroslavBulatov I've added references (Katsurada -- I missed an 'a') to the answer. Also, In the body of the problem you added a plot that indicates a divergence from the asymptotic form. The plot is wrong, and I suspect it is because of insufficient precision in the brute-force summation. I chose an upper limit of 10000, and 42 digits precision for \$\zeta(n)/n! I get the the following triples (n, brute-force,asymptotic): (20,63.503,63.003), (40,154.23,153.73),(60,255.43,243.93),(80,363.42, 362.92). There is no significant deviation. Jul 7, 2021 at 16:08

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


notebook

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.