I am working on a proof that require a closed form (if that is not possible then at least a tight lower bound) of the expression below:
$$A(n,k)=\sum_{i=1}^k \left(1-d^{-i}\right)^{n-1}\left(\prod_{j=1,j \neq i}^k\left( 1- d^{j - i}\right) \right)^{-1}$$
I need to get a bound in terms of $n, k$ and $d$. Here $n, k$ and $d$ are non-negative integers and $d > 0$.
This is not a solution, but rather an ansatz. Several observations with Mathematica suggest that, for $m = k(k+1)/2$, we have
$$ A(n,k) = 1 + \frac{(-1)^{n+k}}{d^{nk-m}} + \cdots + \frac{c_{n,k,j}}{d^{j}} + \cdots + \frac{\binom{n-1}{k+1}}{d^{m+1}} - \frac{\binom{n-1}{k}}{d^{m}} \quad \text{for } 0 < k < n $$
for some integers $c_{n,k,j} \in \Bbb{Z}$ (where $m < j < nk-m$) and
$$ A(n, k) = 1 \quad \text{for } k \geq n. $$
Determining $c_{n,k,j}$ seems not easy, but the observed pattern suggests that it may be written as a combination of binomial coefficients.
Also, its numerical behavior is as follows:
First if $\displaystyle B_{n,k,i}=\frac{(d^i-1)^{n-1}d^{i(k-n)}}{\prod_{j\not =i, 1\leq j\leq k}(d^i-d^j)}$, you have: $\displaystyle A(n,k)=\sum_{i=1}^kB_{n,k,i}$. Put $\displaystyle F_n(x)=\frac{(x-1)^{n-1}x^{k-n}}{\prod_{j=1}^k (x-d^j)}$ and $\displaystyle P_k(x)=\prod_{j=1}^k (x-d^j)$.
A) Suppose that $k\geq n$. Then the degree of the polynomial $(x-1)^{n-1}x^{k-n}$ is $k-1$. We have:
$$F_n(x)=\sum_{i=1}^k\frac{\omega_{n,i,k}}{x-d^i}$$ and it is easy to see that $\omega_{n,i,k}=B_{n,k,i}$. We multiply this equality by $x$ and let $x$ to $\infty$, we get $1=\sum_{i=1}^k B_{n,i,k}=A(n,k)$.
B) Suppose now that $n\geq k+1$. We have $$F_n(x)=\frac{(x-1)^{n-1}}{x^{n-k}P_k(x)}$$
$$F_n(x)=\sum_{m=1}^{n-k}\frac{c_m}{x^m}+\sum_{i=1}^k \frac{B_{n,k,i}}{x-d^i}$$
Again, we multiply by $x$ and let $x\to +\infty$.
We get $\displaystyle 1=c_1+A(n,k)$. Now if $0<R<d$, you have, with $\gamma_R$ the circle $C(0,R)$:
$$c_1=\frac{1}{2i\pi}\int_{\gamma_R} F_n(z)dz$$ and you can bound $|F_n(z)|$ on $C(0,R)$, this gives you a bound for $|A(n,k)|$. Perhaps not simple...
