# Prove that $\exp(r)≥\sum_{k=0}^n\left(\frac{r^k}{k!} \right)$ for any $n∈\mathbb{N}$ and $r≥0$.

Fix $n∈\mathbb{N}$. Prove that $$\exp(r)≥\sum_{k=0}^n\left(\frac{r^k}{k!} \right)$$ for any $n∈\mathbb{N}$ and $r≥0$.

(Hint: you can show that $\exp(r)$ is some kind of upper bound using the fact that a sequence is said to be bounded above if there is some $B∈\mathbb{R}$ such that $s_n≤B$ for all $n∈\mathbb{N}$).

I'm not sure how to prove this inequality, I'm pretty sure that $\sum_{k=0}^{\infty}\left(\frac{r^k}{k!} \right)$ is the power series representation of $\exp(r)$ so how can I show $\exp(r)$ is greater than $\sum_{k=0}^\infty \left(\frac{r^k}{k!} \right)$ I can see from the hint it says to use $\exp(r)$ as an upper bound I'm just not sure how to show this algebraically.

Any help would be greatly appreciated.

Thank you,

• All the terms in the series are positive, so truncating it makes it smaller. – herb steinberg Mar 25 '18 at 21:00

Using the Taylor's theorem with Lagrange's remainder we have that $\exists c \in (0,r)$ so that $$\exp(r)=\sum_{k=0}^{n} \frac{r^k}{k!}+\frac{\exp^{(n+1)}(c)}{(n+1)!}r^{n+1}$$ But $\exp^{(n+1)}(c)=\exp(c) \geq 0$, so we have that $$\frac{\exp^{(n+1)}(c)}{(n+1)!}r^{n+1} \geq 0$$ So: $$\exp(r) = \sum_{k=0}^{n} \frac{r^k}{k!} + \frac{\exp^{(n+1)}(c)}{(n+1)!}r^{n+1}$$ $$\exp(r) - \sum_{k=0}^{n} \frac{r^k}{k!} = \frac{\exp^{(n+1)}(c)}{(n+1)!}r^{n+1} \geq 0$$ So: $$\exp(r) - \sum_{k=0}^{n} \frac{r^k}{k!} \geq 0$$ $$\exp(r) \geq \sum_{k=0}^{n} \frac{r^k}{k!}$$
• Thank you for answering, I haven't seen Lagrange's remainder theorem before so I'm having a hard time seeing how to go from $$\exp(r)=\sum_{k=0}^{n} \frac{r^k}{k!}+\frac{\exp^{(n+1)}(c)}{(n+1)!}r^{n+1}$$ to $$\exp(r)\geq \sum_{k=0}^{n} \frac{r^k}{k!}$$ – jack Mar 25 '18 at 21:23
Simply because the remainder ($\sum\limits_{n + 1}^\infty \cdots$) is a sum of non-negative terms (for $0 \le r$), so it is non-negative itself.