<s>[INST] 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.
 [/INST] 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: $$q_k(x)\geq\frac1k\left(\frac{1}{1+x^2}-\frac{x(1-x^2)}{(1+x^2)^2}-\frac{kx^2} {(1+x^2)^2}\right)$$
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