Asymptotic expansion of $\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $ 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.
 A: 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
A: 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: $$q_k(x)\geq\frac1k\left(\frac{1}{1+x^2}-\frac{x(1-x^2)}{(1+x^2)^2}-\frac{kx^2} {(1+x^2)^2}\right)$$
A: 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 ?
