Simplifying the inequality $nx(1-x)^n \leq \epsilon$ to $n > N(x, \epsilon)$?

Question

How can I reduce the following inequality on $x \in [0, 1]$, and $\epsilon > 0$: $$nx(1-x)^n \leq \epsilon$$ to the this form $$n > N(x, \epsilon)\quad?$$ We can assume $n = 1, 2, 3, \dots$.

My attempt:

$$\ln\big(nx(1-x)^n\big) \leq \ln \epsilon$$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \implies \ln n + n\ln(1-x) \leq \ln\frac{\epsilon}{x}$,

and I am stuck. Since $\ln(1-x) < 0$, I think I must divide by it to change the direction of inequality. But I don't know how to simplify/group $n$ and $\ln n$ while decoupling them from $x$ term.

Ultimately, I want to prove that $\forall n \geq N(x, \epsilon), \quad nx(1-x)^n \leq \epsilon$. Perhaps, I do not need to solve for $n$?

Answer 1

As a real, the solution of $$nx(1-x)^n = \epsilon$$ is given in terms of Lambert function $$n=\frac{1}{\log (1-x)}W\left(\frac{ \log (1-x)}{x}\epsilon \right)$$

Answer 2

Viewing your last sentence, you can procceed as follows.

Take $\varepsilon>0$ (WLOG, you can assume that $\varepsilon<1$).

for $x=0,1$, it is clear that $nx(1-x)^n=0<\varepsilon$ for all $n$.

For $0<x<1$, since $n(1-x)^n\to 0$ ($n\to\infty$) (recall that $0<1-x<1$) you can find $N=N(x,\varepsilon)$ such that if $n\ge N$, $n(1-x)^n<\varepsilon$.

Now $nx(1-x)^n\le n(1-x)^n<\varepsilon$, since $x\le 1$.