By expanding $e^x$ into a series prove the following inequality

Question

By expanding $e^x$ into a series $\sum e^x$ prove that $$\forall x \in \mathbb{R}, x \ge 0 \implies e^{x-1} \ge x$$

Also show when this inequality becomes equality.

I'm not really sure how to attack this problem at all, any tips would be very welcome.

Answer 1

Series expansion is not really needed. Since $e^x$ is a convex function, its graphics lies above any tangent line, so, by considering the tangent line in $x=0$, we have $e^x\geq x+1$ for any $x\in\mathbb{R}$, with equality only in $x=0$, or $$ e^{x-1}\geq x $$ with equality only in $x=1$, as wanted (just replace $x$ with $x-1$).

Answer 2

Set

$y = x - 1; \tag{1}$

then

$x = y + 1, \tag{2}$

so that the inequality

$e^{x - 1} \ge x \tag{3}$

holds if and only if

$e^y \ge 1 + y. \tag{4}$

The inequality (4) is demonstrated in my answer to this question; then (3) immediately follows. Strict inequality when $y \ne 0$ is also therein addressed. QED.

Notes:

(1.) So how does the power series

$e^x = \sum_0^\infty \dfrac {x^n}{n!} \tag{5}$

enter into this discussion? Well, inspection of the proof presented in the the linked citing shows that it rests solidly upon the fact that

$(e^x)' = \dfrac{de^x}{dx} = e^x; \tag{6}$

but (5) readily implies (6), via term-by-term differentiation; indeed, we see from (5) that

$(e^x)' = \sum_0^\infty (\dfrac{x^n}{n!})' = \sum_0^\infty \dfrac{nx^{n -1}}{n!} = \sum_1^\infty \dfrac{x^{n -1}}{(n - 1)!} = \sum_0^\infty \dfrac {x^n}{n!} = e^x. \tag{7}$

Then the only issue which remains is verification that term-by-term differentiation is valid for the series (5); but this follows since the exponential series (5) is both absolutely and uniformly convergent on any closed interval $[-M, M]$ for $) \le M \in \Bbb R$. Details may be found here in gower's blog.

(2.) It may be asked if there isn't a proof of (3)-(4) which more directly uses the power series (5). In response, I would like to say that any demonstration of (3)-(4) from (5) is certain to require some kind of fiddling around with the power series, especially for $y < 0$ (the result is trivial for positive $y$). So I figure my little fiddle tune is as good as almost any, even if it doesn't qualify for first violin. But then again, maybe it does! ;-) End of Notes.