I've been working for a while on these 2 sequences, which arose in cryptography :
$$\forall b\geq2,$$ $$\delta_b = \lim\limits_{p\in\mathbb{N}\to\infty}\left[\sum_{n=0}^{\infty}\left(1-b^{-p}\right)^{b^n}\text{ }-p\right]$$
$$\sigma_b = \lim\limits_{p\in\mathbb{N}\to\infty}\left[\frac{b}{1-b}b^{-p}\left(\sum_{n=b^{p-1}}^{b^p-1}\sum_{k=2}^{n-1}\left(1-b^{-k}\right)^n-\sum_{n=b^{p-1}}^{b^p-1} n\left(1-b^{-n}\right)^n\right)\text{ }-p\right]$$
The original expressions of $\delta_b$ and $\sigma_b$ as they initially appeared in cryptogtaphy were much more nasty looking, and I've already put in quite a lot of work to get to those "nicer" expressions. But now, I don't really know how to move forward...
My ideal objective would be to find a closed form expression for each of these 2 sequences : either we can evaluates the series/sums inside the limit and then set $p\to\infty$, or find a way to work directly on the limits.
If that seems out of reach, I have a secondary objective : finding $\lim\limits_{b\to\infty} \delta_b$ and $\lim\limits_{b\to\infty} \sigma_b$.
Maybe we can switch the order of the limits here, but it still is complicated. From numerical simulation, $\delta_b$ is increasing, $\sigma_b$ is decreasing, and I actually suspect those sequences converge to the same limit $\frac{1}{e}$. Now I need to prove it...
Any improvement in this problem would be of great help, and with concrete applications in cryptography.
Any suggestions ?