One version of Bernouilli's inequality states that
$$(1 + x)^n > 1 + nx $$
for any real number $ x > -1$ with $x \neq 0$ and integer $ n \geq 2$.
I was just reading this proof on Wolfram Mathworld and had some trouble with a key step for the case of $x < 0$. After defining $y \equiv -x$, the proof starts by writing
$$(1 - y)^n = 1 - yn + \frac{1}{2}n(n-1)y^2 - \frac{1}{6}n(n-1)(n-2)y^3 + . . .$$
It is then argued that
Since each power of $y$ multiplies by a number $< 1$ and since the absolute value of the coefficient of each subsequent term is smaller than the last, it follows that the sum of the third order and subsequent terms is a positive number.
Unfortunately, I am having trouble seeing how "the absolute value of the coefficient of each subsequent term is smaller than the last". For if $n$ is sufficiently large, then it seems that the coefficients will initially increase in size (in absolute value).
Am I missing something obvious here? If not, is there a simple way to fix the argument?
let $f(x)=(1+x)^n-1-nx,$ where $x>-1$ and $n\geq2$.
Thus, $f'(x)=n(1+x)^{n-1}-n,$ which gives the $x_{min}=0$ and for $x\neq0$ we obtain $f(x)>0.$