TITLE: Asymptotic expansion of $\sum_{k=1}^n {n \choose k} (-1)^k \frac 1 {1-x^k} $
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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