Show that $e^x > 1 + x + x^2/2! + \cdots + x^n/n!$ for $n \geq 0$, $x > 0$ by induction

Show that if $n \geq 0$ and $x>0$, then

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

Not sure where to get started with this induction proof.

Solution 1:

Base case: $e^x > 1$ for $x > 0$.

Induction: suppose we are given $k$ such that for all $x > 0$ $$ e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!} $$ Change variables to $t$ and integrate both sides: $$ \int_0^x e^t \; dt > \int_0^x \left( 1 + t + \frac{t^2}{2!} + \cdots + \frac{t^k}{k!} \right) \; dt $$ $$ e^x - 1 > x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!} $$ $$ e^x > 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^{k+1}}{(k+1)!} $$

Note that if $f > g$ on $(a,b)$ and both are integrable, then $\int_a^b f > \int_a^b g$ (not just $\ge$).

Solution 2:

It rather depends on your definition of $e^x$.

Suppose you know $\frac{d}{dx} e^x = e^x$ and $e^0=1$.

Call $f_n(x) = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$. Then your inductive step could say $f_{n+1}(0) = 1$ and $\frac{d}{dx} f_{n+1}(x) = f_n(x) \lt e^x$ for $x \gt 0$ so $f_{n+1}(x) \lt e^x$.