Prove $\forall x\in \mathbb{R} ^{+}$ we have $e^{x}\geq \dfrac {x^{n+1}} {\left( n+1\right) !}$.

Question

Prove the following statement:

$$\forall x\in \mathbb{R} ^{+}\,:\,\, e^{x}\geq \dfrac {x^{n+1}} {\left( n+1\right) !}$$

Answer 1

$$e^x>e^x-1 = \sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)!}$$ This series contains $\frac{x^{n+1}}{(n+1)!}$ for all $n \in \mathbb{N}$, and each term is positive. The OP's claim follows by taking only one term in this series, which is trivially less than all of them added together.

Answer 2

Use the Taylor series: $$ e^{x}=\sum_{n=0}^\infty\dfrac {x^{n}} {\left( n\right) !}.$$

And show that $e^x$ must necessarily be greater than or equal to any finite approximation in this manner.