exponential function and mathematical induction

Question

May I ask how to solve the problem? Use mathematical induction to prove that for $x\geq0$ and positive integer $n$, $$e^x \geq 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}$$

Answer 1

Well, first off you need $x >0$, not just $ x \geq 0$. So, you know $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, which is, by definition, $\lim_\limits {k \to \infty} \sum_{n=0}^k\frac{x^n}{n!}$. Let $a_n = \frac{x^n}{n!}$, and $b_k = \sum_{n=0}^k\frac{x^n}{n!} = \sum_{n=0}^k a_n$. Clearly, $a_n > 0$, since $x>0$.

Moreover, $b_{k+1} = b_k + a_{k+1} > b_k$. So, $(b_k)_{k \in \mathbb{N}}$ is an increasing sequence, and it converges, it converges to $\sup( \{ b_k \})$. So, $b_k \leq e^x$, for all $k$. But, then we have $b_k < b_{k+1} \leq e^x$, giving $b_k < e^x$, for all k.

I didn't use induction and I'm not sure if this is the best answer. But, here you are.

Answer 2

Integrate both sides from $0$ to $x$ and add $1$.