Is it true that for $n\in\mathbb{N}$ and $x> 0$ we have $$ e^x-\sum_{k=0}^n \frac{x^k}{k!} \leq e^x\frac{x}{n}? $$
What I am getting is this: Use Taylor's series with remainder. We know that $$ e^\lambda = \sum_{k=0}^r \frac{\lambda^k}{k!} + \frac{e^{c\lambda}\lambda^{r+1}}{(r+1)!}, $$ for some $c \in [0,1]$. Therefore $$ \frac{\lambda^{r+1}}{(r+1)!} \leq e^\lambda - \sum_{k=0}^r \frac{\lambda^k}{k!} \leq e^\lambda \frac{\lambda^{r+1}}{(r+1)!}. $$
We can get a different upper bound by comparison to a geometric series, when $\lambda < r+2$: $$ \sum_{k=r+1}^\infty \frac{\lambda^k}{k!} \leq \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \left(\frac{\lambda}{r+2}\right)^t = \frac{\lambda^{r+1}}{(r+1)!} \frac{r+2}{r+2-\lambda}. $$
The left side is $\sum\limits_{k=n+1}^{\infty} \frac {x^{k}} {k!}$. Note that $k! >(k-1)! n$ for $k >n$. Hence an upper bound for LHS is $\frac 1 n\sum\limits_{k=n+1}^{\infty} \frac {x^{k}} {(k-1)!}$ which is $\frac x n\sum\limits_{k=n+1}^{\infty} \frac {x^{k-1}} {(k-1)!} \leq \frac x n e^{x}$.