Prove $e^x > 1 + x +\frac{ x^2}{2!} + ··· + \frac{x^n}{n!}$ for $x>0$ with induction.

Question

I'm having trouble starting with the base case even. So far I've set $$ f_n(x)=1 + x +\frac{ x^2}{2!} + ··· + \frac{x^n}{n!}. $$ For $n = 0$, $e^x>f_0(x)$, but I'm not sure that's the correct base case since usually $n=1$ is used.

From there, I'm having dificulty establishing $e^x>f_{k+1}(x)$.

I can use derivation, but cannot use integration of any kind. I must use indiction. I do not have a formal definition of $e^x$, and my textbook is not useful or good.

EDIT: I cannot use the power series of $e^x,$ my apologies for taking the question incorrectly!

Answer 1

If you know calculus, we can consider the Maclaurin series for $e^x$: $$ e^x = 1+x+\frac1{2!}x^2+\frac1{3!}x^3+\frac1{4!}x^4+\cdots.$$ Notice that each term in the series is positive! The function in the inequality is just this expression, but terminiating after a finite number of terms. But there are always more terms after where you terminated, and these terms are positive. Thus $e^x$ is greater than the expression you gave.

Answer 2

By integration by parts $$ e^x-f_n(x) = \int_{0}^{x}\frac{(x-t)^n}{n!}e^t\,dt $$ and the integral of a positive function over an interval is positive. Actually

$$ e^x-f_n(x) = \frac{x^{n+1}}{n!}\int_{0}^{1}(1-t)^n e^{xt}\,dt = \frac{x^{n+1}e^x}{n!}\int_{0}^{1}t^n e^{-xt}\,dt$$ is greater than $$ \frac{x^{n+1}e^x}{n!}\int_{0}^{1}t^n(1-t)^x\,dt =\frac{x^{n+1}e^x}{n!}\cdot\frac{\Gamma(n+1)\Gamma(x+1)}{\Gamma(n+x+2)}=\frac{x^{n+1}e^x}{(x+1)_{n+1}}.$$