Lower bound for $(1+x)^n$ I am trying to prove the following bound for $n \in N$ and $x \in [-1,0)$, but x closer to 0:
$ e^{nx} - \frac{1}{2}nx^{2}e^{(n-1)x} \leq (1+x)^n $
I have:
$ e^{nx} - \frac{1}{2}nx^{2}e^{(n-1)x} = (1 - \frac{1}{2}nx^{2}e^{-x})e^{nx} \leq 1 - \frac{1}{2}nx^{2}e^{-x} \leq 1 - x(\frac{nx}{2}) $
And looking at some limit representation for $e$ or some $e$ Taylor expansion but I am stuck.
 A: For $x\in [-1,0)$, we have
$$(1-x)^n\geq 1^n=1$$
and (as you suggested)
$$e^{nx}-\tfrac{1}{2}nx^2e^{(n-1)x}\leq 1-\frac{1}{2}nx^2e^{-x}\leq 1.$$
The result immediately follows.
A: Fisrt, lets make $x = x_n < 0$ such that $1 - \frac{1}{2}nx^2e^{-x} > 0$ then consider the following taylor expansions:
$(1+x)^n = 1 + xn + \frac{1}{2}nx^2 + (n^3 - 3n^2 -2n)\frac{x^3}{3} + O(x^4)$
$e^{nx} - \frac{1}{2}nx^2e^{(n-1)x} = 1 + xn + \frac{1}{2}nx^2 + (2n^3 - 6n^2 + 6n)\frac{x^3}{3} + O(x^4)$
Substracting them we obtain:
$(1+x)^n - (e^{nx} - \frac{1}{2}nx^2e^{(n-1)x}) = (n^3 - 3n^2 -2n)\frac{x^3}{3} - (2n^3 - 6n^2 + 6n)\frac{x^3}{3} + O(x^4)$
We just have to proove that
$(n^3 - 3n^2 -2n)\frac{x^3}{3} - (2n^3 - 6n^2 + 6n)\frac{x^3}{3} \geq 0$
i.e.
$(n^3 - 3n^2 -2n)\frac{x^3}{3} \geq (2n^3 - 6n^2 + 6n)\frac{x^3}{3}$
i.e.
$(n^3 - 3n^2 -2n) \leq (2n^3 - 6n^2 + 6n) \iff (n^3 - 3n^2 + 8n) \geq 0 $
It should be trivial that for $n \geq 0$ the cubic polynomial images are above $0$, and in $0$ it is exactly $0$.
$\blacksquare$
