Lower bound for $(1+x)^n$

Question

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.

Answer 1

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.

Answer 2

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 $0$. $\blacksquare$