Asymptotic growth of $\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}\binom{n}{k}}{1-p^{k}}$ as $n \to \infty$

Question

This is the expected value of the height of a skip list with $n$ elements. I have a formula for this value, but it is defined by a summation. It is, where $n$ is the number of elements, $$f_p(n) =\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}\binom{n}{k}}{1-p^{k}}$$ I want to get a closed form for this value, in terms of $n$. I have tried using Pascal's identity, but have gotten nowhere. The first $100$ values of the numerators and denominators of this for $p = 1/2$ can be found under the OEIS, respectively A158466 and A158467. Is it possible to get a closed formula for this value and if so, what is it?

Edit: About two years later, I am still interested in the problem, but I doubt there is a closed formula. Would it be possible to find the asymptotic growth instead?

From the Wikipedia page on skip lists, the growth rate is $O(\ln(n))$, but what is the exact value of $$\lim_{n \to \infty}\frac{f_p(n)}{\ln(n)}$$

as well as the additional constant (i.e. what are the values of $c_1, c_2$ such that $f_p(n) = c_1\ln(n) + c_2 + o(1)$)? Further expansion would also be appreciated.

Looking at the graphs, it seems that $f_p(n) \sim -\log_p(n)$, which means $c_1 = - \frac{1}{\ln(p)}=-\log_p(e)$, but I couldn't find a value for $c_2$ or prove that that is the correct value for $c_1$.

Why do you think there is a closed form? I doubt there is one. — GEdgar, Sep 20 '18 at 00:17
I don't know if there is a closed form. I am wondering if there is, and if so, what it is. — Varun Vejalla, Sep 20 '18 at 00:20
It's equivalent to $\enspace\displaystyle \sum\limits_{v=0}^\infty \left(1-\left(1-\frac{1}{2^v}\right)^n\right),$ . No "closed" form. Your sum is something like a closed form for this series for natural numbers $n$. — user90369, Sep 20 '18 at 11:33
It's equivalent to $\sum _{k=1}^n \frac{(-2)^k \binom{n}{k}}{-1+2^k}=\underset{x\to 1}{\text{lim}}\mathcal{M}_s^{-1}\left\left(-1+\left(1-2^s\right)^n\right) \pi \cot (\pi s)\right$ where $ {M}_s^{-1}$ is Inverse Mellin Transfrom. — Mariusz Iwaniuk, Sep 20 '18 at 11:58
@metamorphy Awesome, that answer of yours seems to address my question pretty well. — Varun Vejalla, Aug 14 '20 at 05:55
I've removed my comment (it had a mistake) and posted a fixed version as an answer. — metamorphy, Aug 14 '20 at 06:31

metamorphy · Accepted Answer · 2020-08-14T06:36:54.127

An alternative expression for $f_p(n)$ is obtained in my later answer: $$f_p(n)=\frac12-\frac{H_n}{\ln p}+\frac{1}{\pi}\Im\sum_{m=1}^\infty\frac1m\prod_{k=1}^n\left(1+\frac{2m\pi i}{k\ln p}\right)^{-1},\quad H_n=\sum_{k=1}^n\frac1k.$$ This confirms the value of $c_1=-1/\ln p$. But, as for the "remainder", $$\lim_{n\to\infty}e^{z\ln n}\prod_{k=1}^n\left(1+\frac{z}{k}\right)^{-1}=e^{-\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)^{-1}e^{z/k}=\Gamma(1+z),$$ so that the absolute value of each term of the $\sum_{m=1}^\infty$ tends to a finite limit as $n\to\infty$ and, while these limits are exponentially decaying with $m$, the terms themselves are oscillating like $\exp(\ldots i\ln n)$.

Thus, the "remainder" $f_p(n)+\ln n/\ln p$ oscillates around $1/2-\gamma/\ln p$, and there's no $c_2$.

Huh, that's unexpected - I had expected there to be a $c_2$. Thanks for proving my guess wrong! — Varun Vejalla, Aug 23 '20 at 06:32

Asymptotic growth of $\sum_{k=1}^{n}\frac{\left(-1\right)^{k+1}\binom{n}{k}}{1-p^{k}}$ as $n \to \infty$

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