I'm working with a sequence of $k$ by $k$ matrices $M^n$ whose entries satisfy
$$M^n_{ij} = \binom{j-1}{i-1} \sum_{l=0}^{j-i} \binom{j-i}{l} (-1)^{l} (\frac{1}{i+l})^n, \quad n \in \mathbb{N}$$
I've been trying without luck to simplify this sum, specifically
$$ \sum_{l=0}^{j-i} \binom{j-i}{l} (-1)^{l} (\frac{1}{i+l})^n $$
One avenue is to expand out the binomials in the denominator:
$$\sum_{l=0}^{j-i}\frac{\binom{j-i}{l} (-1)^l}{\sum_r^n \binom{n}{r}i^nl^{r-n}} $$
But I think this may be going backwards rather than forwards.
I also tried the strategy in https://math.stackexchange.com/a/3369814/978915 of using generating functions, i.e.
$$\bigg(\frac{1}{l+i}\bigg)^n = n![x^n]\exp\big(\frac{x}{l+i}\big)$$
where $[]$ is the "coefficient of" operator. This leads to
$$\sum_{l=0}^{j-i} \binom{j-i}{l} (-1)^{l} \big(\frac{1}{i+l}\big)^n = n![x^n] \sum_{l=0}^{j-i} \binom{j-i}{l} (-1)^{l} \exp\big(\frac{x}{l+i}\big) $$
$$= n![x^n] \sum_{l=0}^{j-i} \binom{j-i}{l} \exp(-\mathbf{i}\pi l) \exp\big(\frac{x}{l+i}\big) = n![x^n] \sum_{l=0}^{j-i} \binom{j-i}{l} \exp\big(\frac{\mathbf{i}\pi l(l+i) + x}{l+i}\big) $$
Unfortunately the $(-1)^l$ and the exponential can't be simplified in the same way as in that example.
There's also this interesting identity from https://math.stackexchange.com/a/3657580/978915 , which does not cover this case because here $0 \leq \tau \leq k$
