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. I would like to obtain a solution of that inequality, not a proof of existence of such $n$ and $k$.
It is possible to show that the inequality in the first place may be turned into this simplified one (for even $n$):
$$ \sum_{l=1}^{n} \frac{q^l - p^l}{l!} > \frac{1}{k} $$
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.
Let $m=\lceil -\mathrm ep\rceil$. Then the ratio of magnitudes of successive terms beyond $m$ is less than $1/\mathrm e$. Now choose $n$ such that
$$ \frac{|p|^m}{m!}\frac1{1-1/\mathrm e}\mathrm e^{-(n-m)}\le\frac12\mathrm e^p\;. $$
Since the $m$-th term has magnitude $|p|^m/m!$ and the magnitudes of the remaining terms are bounded by a geometric sequence with ratio $\mathrm e$, the terms beyond $n$ can add up to at most the left-hand side, and if this is bounded by the right-hand side, the sum up to $n$ must be positive, since the series converges to $\mathrm e^p$. Thus,
$$ n=\left\lceil m-p+\log\left(\frac2{1-1/\mathrm e}\frac{|p|^m}{m!}\right)\right\rceil $$
ensures that
$$ \sum_{l=0}^{n} \frac{p^l}{l!} $$
is positive.