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 $\frac{(n-1)}{2}x>\alpha-1.$


Well, of course. Equations in $x$ of the form $p^x=ax+b$ have solutions given by the Lambert-$W$ function. In your specific case, we have

$$(1+x)^n=(\alpha x)n+1$$

And the corresponding solution in $n^*$ is

$$n^*=-\frac{W_{-1}\left(-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)}{\ln(1+x)}-\frac1{\alpha x}$$

Hence, for all $n\geq n^*$ we have $(1+x)^n\geq 1+n\alpha x$.

There's the bound

$$W_{-1}(-e^{-u-1})\geq -1-\sqrt{2u}-u.$$

Setting $-e^{-u-1}=-\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}$, we get

\begin{align} u&=-1-\ln\left(\frac{\ln(1+x)}{\alpha x}\,(1+x)^{-\frac1{\alpha x}}\right)\\ &=-1-\left(\ln\left(\frac{\ln(1+x)}{\alpha x}\right)+\ln\left((1+x)^{-\frac1{\alpha x}}\right)\right)\\ &=-1-\left(\ln(\ln(1+x))-\ln(\alpha x)-\frac1{\alpha x}\ln(1+x)\right)\\ &=\ln(\alpha x)+\frac{\ln(1+x)}{\alpha x}-1-\ln\Big(\ln(1+x)\Big). \end{align}

So, depending on how precise you wish to be, you may use

$$n^*\geq \frac{1+\sqrt{2u}+u}{\ln(1+x)}-\frac1{\alpha x},$$

where $u$ is as above.


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