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I'm trying to find the limit (and asymptotic expansion as $n\to\infty $) of $$\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $$

for $0<x<1$.

So far, I have no idea...

I found this question when dealing with the expected value of $\max (X_1, \dots X_n)$ when $X_1, \dots, X_n$ are geometric random variables with same success probability $p$...

Thus, the expected value is this sum for $x=1-p$ (in absolute value)...

Could you help me please ?

Thanks in advance.

Bastien Tourand
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    Asymptotic expansion when $x\to 1$? – popi Jun 21 '21 at 21:07
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    For $n\to\infty$ see this. – metamorphy Jun 21 '21 at 23:17
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    @ThomasAndrews Your suggestion of writing $S$ as $\sum_{i \geq 0} ((1 - x^i)^n - 1)$ is sufficient to get the $\ln n/\ln x + O(1)$ estimate for fixed $x$ and $n \to \infty$, since the difference between $S$ and $$\int_0^\infty ((1 - x^i)^n - 1) di = \frac {H_n} {\ln x}$$ is bounded. – Maxim Jun 22 '21 at 12:58
  • I think this is a very interesting question, but the format is that of a PSQ, and so it may get closed and even deleted, and those who answer may get penalized. Please try to add some context; where you found the question, if it is from a class or textbook, what are some of the other topics being discussed at the time, etc. – robjohn Jun 22 '21 at 17:20
  • @metamorphy I was expecting a simpler solution than yours and I don't know the tools you are using... So I consider it as incomplete regarding the context of this question... (added context)... – Bastien Tourand Jun 22 '21 at 18:59
  • @Maxim: Since the integrand is monotonically increasing (from $-1$ to $0$), the integral is less than $1$ above the sum. This is the "$-[0,1]$" after the integral in my answer. – robjohn Jun 22 '21 at 19:57
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    The OP added clarification and some context after being requested. I don't think it deserves to be closed for lack of context. – robjohn Jun 25 '21 at 17:15
  • @BastienTourand: even though you have added where this question came from, perhaps you could expand on the parts alluded to, but elided by the ellipses (the parts that follow the "..."s). – robjohn Jun 25 '21 at 17:21
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    Have you read this ? https://math.stackexchange.com/a/26214/312 – leonbloy Jun 27 '21 at 18:21

3 Answers3

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A First Approach $$ \begin{align} \sum_{k=1}^n\binom{n}{k}(-1)^k\frac1{1-x^k} &=-1+\sum_{k=1}^n\binom{n}{k}(-1)^k\frac{x^k}{1-x^k}\tag1\\ &=-1+\sum_{k=1}^n(-1)^k\binom{n}{k}\sum_{j=1}^\infty x^{jk}\tag2\\ &=-1+\sum_{j=1}^\infty\sum_{k=1}^n(-1)^k\binom{n}{k}x^{jk}\tag3\\ &=\sum_{j=0}^\infty\left(\left(1-x^j\right)^n-1\right)\tag4\\ &\sim\int_0^\infty\left(1-\left(1-n^{-t}\right)^n\right)\frac{\log(n)}{\log(x)}\,\mathrm{d}t\tag5 \end{align} $$ Explanation:
$(1)$: $\frac1{1-x^k}=1+\frac{x^k}{1-x^k}$ and $\sum\limits_{k=1}^n\binom{n}{k}(-1)^k=-1$
$(2)$: $\frac{x^k}{1-x^k}=\sum\limits_{j=1}^\infty x^{jk}$
$(3)$: change order of summation
$(4)$: apply the binomial theorem, then note that the $j=0$ term is $-1$
$(5)$: approximate the sum with an integral using $j=-t\frac{\log(n)}{\log(x)}$

Since $\lim\limits_{n\to\infty}\int_0^\infty\left(1-\left(1-n^{-t}\right)^n\right)\,\mathrm{d}t=1$, the sum should be asymptotic to $\frac{\log(n)}{\log(x)}$.

A Second Approach

The sum in $(4)$ can also be approximated by $$ \begin{align} \sum_{j=0}^\infty\left(\left(1-x^j\right)^n-1\right) &=\int_0^\infty\left(\left(1-x^t\right)^n-1\right)\mathrm{d}t-[0,1]_\#\tag6\\ &=\sum_{k=1}^n(-1)^k\binom{n}{k}\int_0^\infty e^{kt\log(x)}\mathrm{d}t-[0,1]_\#\tag7\\ &=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\frac1{k\log(x)}-[0,1]_\#\tag8\\ &=\frac{H_n}{\log(x)}-[0,1]_\#\tag9 \end{align} $$ Explanation:
$(6)$: overestimate the sum with an integral
$\phantom{\text{(6):}}$ the variation of the integrand is equal to $1$
$\phantom{\text{(6):}}$ $[0,1]_\#$ represents a real number between $0$ and $1$
$(7)$: $x^t=e^{t\log(x)}$, then apply the binomial theorem
$(8)$: evaluate the integral
$(9)$: apply this answer

robjohn
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  • Incredibly simple (after reading it !) ! For the case $n= 10$ and $x=0.9$, $\frac{H_n}{\log (x)}\sim -27.7995$ for an exact value $-28.2995$ ! More than impressive. Cheers :-) – Claude Leibovici Jun 23 '21 at 10:29
  • @ClaudeLeibovici: If we use $0.5$ for the $[0,1]$, it is pretty close (not only in your example). – robjohn Jun 23 '21 at 12:56
  • I was just about to make the same comment. In French, we use to say something like "great minds meet" This is fully true except that I am not one. By the way, I made my expansion using unsigned Stirling numbers. I suppose that you noticed what is the first coefficient. Cheers and thanks again :-) – Claude Leibovici Jun 23 '21 at 13:18
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Not a complete answer, but an idea of where to look and some lower bounds.

Let $$p_k(x)=\frac{1-x^k}{1-x}=1+x+\dots+x^{k-1}$$

Then:

$$\frac{1}{1-x^k}=\frac{1}{k(1-x)}+\frac{(k-1)+(k-2)x+\cdots+x^{k-2}}{kp(x)}$$

Also, $$\begin{align}\sum_{k=1}^{n}(-1)^k\binom{n}k\frac1k&=\int_0^1\frac{(1-t)^n-1}{t}\,dt\\ &=\int_0^1 \frac{t^n-1}{1-t}\,dt\\ &=-\int_0^1(1+t+t^2+\cdots+t^{k-1})\,dt\\ &=-\left(1+\frac12+\cdots+\frac{1}{n-1}\right) =-H_{n-1} \end{align}$$

So if $f_n(x)$ is your term, then $$f_n(x)=\frac{-H_n}{1-x}+\sum_{k=1}^{n}(-1)^k\binom nkq_k(x)$$

When $$q_k(x)=\frac{(k-1)+(k-2)x+\cdots+x^{k-2}}{kp(x)}$$

Notice that $$q_k(x)=\frac{x^{k-2}p_k’(1/x)}{kx^{k-1}p_k(1/x)}=\frac{1}{x}\frac{p_k’(1/x)}{kp_k(1/x)}$$

For any polynomial $p$ with no repeating roots, $$\frac{p’(y)}{p(y)}=\sum_j \frac{1}{y-r_i}$$ where the $r_i$ are the roots of the polynomial.

So:

$$\begin{align}q_k(x)&=\frac{1}{k}\sum_{j=1}^{k-1}\frac{1}{1-xe^{2\pi ij/k}}\\ &= \frac{1}{k}\sum_{j=1}^{k-1}\frac{1-xe^{-2\pi ij/k}}{1+x^2-2x\cos(2\pi ij/k)}\\ &=\frac1k\sum_{j=1}^{k-1}\frac{1-x\cos(2\pi j/k)} {1+x^2-2x\cos(2\pi j/k)} \end{align} $$

The last step because we know the imaginary values cancel.

Now:

$$\begin{align}\frac{1}{1+x^2-2x\cos\theta}&=\frac{1}{1+x^2}\sum_{p=0}^{\infty}\left(\frac{2x\cos\theta}{1+x^2}\right)^p\\ &\geq\frac{1}{1+x^2}+\frac{2x\cos\theta}{(1+x^2)^2} \end{align}$$

So:

$$\frac{1-x\cos\theta}{1+x^2-2x\cos\theta}\geq\\ \frac{1}{1+x^2}+\frac{2x\cos\theta}{(1+x^2)^2} -\frac{x(1+x^2)\cos\theta}{(1+x^2)^2}-\frac{2x^2\cos^2\theta}{(1+x^2)^2}\\= \frac{1}{1+x^2}+\frac{x(1-x^2)\cos\theta}{(1+x^2)^2}-\frac{2x^2\cos^2\theta}{(1+x^2)^2} $$

Now:

$$\sum_{j=1}^k\cos(2\pi j/k)=-1$$

and $$\sum_{j=1}^k\cos^{2}(2\pi j/k)=\frac k2$$

So: $$q_k(x)\geq\frac1k\left(\frac{1}{1+x^2}-\frac{x(1-x^2)}{(1+x^2)^2}-\frac{kx^2} {(1+x^2)^2}\right)$$

Thomas Andrews
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If you are looking for the asymptotics of $$f_n=\sum_{k=1}^n (-1)^k\,{n \choose k} \, \frac 1 {1-x^k}$$ for $x \to 1$,we have (from an old work of mine) $$f_n=\frac{\left|S_{n+1}^{(2)}\right|}{n!}\frac 1{(x-1)}+\frac{\left|S_{n+1}^{(2)}\right|-\left|S_{n+1}^{(1)}\right|}{2 \,n!}-\frac{\left|S_{n+1}^{(2)}\right| }{12\, n!}(x-1)+\frac{\left|S_{n+1}^{(2)}\right| }{24\, n!}(x-1)^2+O((x-1)^3)$$ where appear unsigned Stirling numbers of the first and second kinds.

Trying for $n=10$ and $x=\frac 9{10}$, the exact result is $$-\frac{22954528746372225790890814301786200}{811128802013725681512205937948047}$$ while the above truncated expansion gives $$-\frac{19017311}{672000}$$

The absolute difference between these two numbers is $8.32\times 10^{-5}$.

Question

The next term of the expansion is $$-\frac{c_n}{720 \,n!} (x-1)^3$$ where the $c_n$'s form the sequence $$\{45,245,950,5206,33516,248292,2082096,19504944,201944160,\cdots\}$$ What could they be ?

Claude Leibovici
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