While trying to show that partial exponential series evaluated at two different values are strictly increasing provided that sufficient number of terms are applied I stuck at a problem. Given two rational numbers $p,q \in \mathbb{Q}$ with $p<q$, find $n,k \in \mathbb{N}$ s. t. the following holds:
$$ \sum_{l=1}^{n} \frac{q^l - p^l}{l!} - \frac{|q|^{n+1}}{(n+1)!} - \frac{|p|^{n+1}}{(n+1)!} > \frac{1}{k}$$
The first term is the difference between two partial sums and the last two terms indicate the half-bounds in which the partial sums are located after $n$-th step. The best what I expect is a closed-form expression, and the least is an algorithm whose number of steps may be estimated beforehand. I pursue a routine which I might implement in a computer that would calculate those two numbers.
I also think it is possible to show that the inequality may be turned into this simplified one (provided $n$ is even):
$$ \sum_{l=1}^{n} \frac{q^l - p^l}{l!} > \frac{1}{k} $$
It follows from the fact that for even $n$:
$$ \sum_{l=0}^{n-g-1} \frac{p^l}{l!} > \sum_{l=0}^{n-g-3} \frac{p^l}{l!} $$
with $n \geq |p| + g + 1$. It is done by subtracting maximal values of bottom margins in consecutive terms.
For the sake of simplicity consider a function $e_n(x)=\sum_{l=0}^{n} \frac{x^l}{l!}$. By Taylor formula in Lagrange form, there exist numbers $0<\theta_p,\theta_q<1$ such that $$e^p-e_n(p)= \frac{e^{p\theta_p }p^{n+1}}{(n+1)!}$$ and $$e^q-e_n(q)=\frac{e^{q\theta_q}q^{n+1}}{(n+1)!}.$$
Put $\Delta=e^q-e^p$, pick $1/k\le\Delta/3$, and $n$ such that both $\frac{|p|^{n+1}}{(n+1)!}$ and $(e^q+1)\frac{|q|^{n+1}}{(n+1)!}$ are not greater than $\Delta/3$ (here you can use Stirling approximation for $n!$, climing that there exists a number $0<\theta(n)<1$ such that $n!=\sqrt{2\pi n}\left(\frac ne\right)^n e^{\frac{\theta(n)}{12n}}$). Then $$ e_n(q)-e_n(p) - \frac{|q|^{n+1}}{(n+1)!} - \frac{|p|^{n+1}}{(n+1)!} =$$ $$ e^q-e^p-\frac{e^{q\theta_q}q^{n+1}}{(n+1)!}+ \frac{e^{p\theta_p}p^{n+1}}{(n+1)!}- \frac{|q|^{n+1}}{(n+1)!} - \frac{|p|^{n+1}}{(n+1)!}>$$ $$\Delta- (e^q+1)\frac{|q|^{n+1}}{(n+1)!}- \frac{|p|^{n+1}}{(n+1)!}\ge\frac{\Delta}{3}\ge \frac{1}{k}.$$