Show that an inequality is true using mathematical induction and the mean value theorem

Question

A question in my math book is: "Use mathematical induction to show that:

$$e^x>1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}$$

if $x>0$ and $n$ is any positive integer."

Apparently the solution involves applying the Mean Value Theorem on:

$$g(t)=e^t-1-t-\frac{t^2}{2!}-...-\frac{t^{k+1}}{(k+1)!}$$

Then:

$$\frac{g(x)}{x}=\frac{g(x)-g(0)}{x-0}=g'(c)$$

Then it states that $g'(c)$ is positive.

QUESTION 1: why is $g'(c)$ positive?

QUESTION 2: how would one answer (in small understandable steps) the original question of my math book? I don't fully understand the method used.

Thanks!

Answer 1

Let $P_n(x)=1+x+\dfrac{x^2}{2!}+...+\dfrac{x^n}{n!}$.

*Initialisation:

$\mathrm e^x >1$ is true for x >0, since the exponential function is increasing and $\mathrm e^0=1$.

Induction step:

We'll use this corollary of the Mean value theorem:

Let $f$ and $g$ be continuous functions on an interval $I$, differentiable in the interior of $I$, and $x_0\in I$. Suppose $f(x_0)\ge g(x_0)\;$ and $\;f'(x)>g'(x)$ for all $x>x_0\;$ in $I$.

Then $\;f(x)> g(x)$ for all $x>x_0\;$ (in $I$).

Now suppose $\mathrm e^x>P_n(x)$ for all $x>0$, and observe $(\mathrm e^x)'=\mathrm e^x$, $\, (P_{n+1}(x))'=P_n(x)$. So you just have to apply the corollary.

Answer 2

Claim $(n)$:

$$\qquad\forall x>0,\qquad e^{x} > \sum_{k=0}^{n}\frac{x^k}{k!}$$

Case $n=0$: trivial. Induction step: given Claim $(n)$, it follows that:

$$ e^{x} = 1+\int_{0}^{x}e^{z}\,dz > 1+\int_{0}^{x}\sum_{k=0}^{n}\frac{z^{n}}{n!}\,dz = \sum_{k=0}^{n+1}\frac{x^k}{k!}.$$