A stronger form of Bernoulli's inequality Let $x\geq 0$ be fixed. By Bernoulli's inequality we know that for all $n \in \mathbb{N}_0$,
$$
(1+x)^n \geq 1+nx
$$
Now, let $\alpha \geq 0$ be fixed as well. By a limit argument, we know that
$$
(1+x)^n \geq 1+\alpha nx
$$
for large $n$. But how large does $n$ have to be? More specifically, can we find some expression for $N \in \mathbb{N}$ depending on $x$ and $\alpha$ such that the inequality holds for all $n \geq N$?
If it helps, $N$ does not necessarily have to be optimal. I am specifically interested in the case where $\alpha \in \mathbb{N}$, and even more specifically in the case $\alpha=2$.

If $n\ge 2$ then $$(1+x)^n\ge 1+nx+\frac{n(n-1)}{2}x^2.$$So you just need $\frac{(n-1)}{2}x>\alpha-1.$
