Approximations for the partial sums of exponential series

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 $$ \frac{\lambda^{r+1}}{(r+1)!} \leq e^\lambda - \sum_{k=0}^r \frac{\lambda^k}{k!} \leq e^\lambda \frac{\lambda^{r+1}}{(r+1)!}. $$ You can also get these estimates using more elementary means: $$ \sum_{k=r+1}^\infty \frac{\lambda^k}{k!} = \frac{\lambda^{r+1}}{(r+1)!} \left[ 1 + \frac{\lambda}{r+2} + \frac{\lambda^2}{(r+2)(r+3)} + \cdots \right] < \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \frac{\lambda^t}{t!} = \frac{\lambda^{r+1}}{(r+1)!} e^\lambda. $$ We can get a different upper bound by comparison to a geometric series, when $\lambda < r+2$: $$ \sum_{k=r+1}^\infty \frac{\lambda^k}{k!} \leq \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \left(\frac{\lambda}{r+2}\right)^t = \frac{\lambda^{r+1}}{(r+1)!} \frac{r+2}{r+2-\lambda}. $$