Finding the smallest $n$ such that $\sqrt{n}(1 - \delta^2)^n < \epsilon$?

Question

I am trying to solve for a lower bound $N_{\epsilon, \delta}$ on $n$ such that $$n \geq N_{\epsilon, \delta} \implies \sqrt{n}(1 - \delta^2)^n < \epsilon$$ holds, where $\epsilon > 0$ and $0 < \delta < 1$. I've tried applying the inequality $1 + nx \leq e^{nx}$, and using logarithms, but I can't figure this out. Could someone please give me a hint?

Thanks!

Answer 1

We have that

$$\epsilon > \sqrt{n}(1-\delta^2)^n > (1-\delta^2)^n$$

which gives us

$$\log \epsilon > n\log(1-\delta^2) \implies n > \frac{\log\epsilon}{\log(1-\delta^2)}$$

The inequality flips after the division because the $\log$'s are negative.