If we have $a>1$ and $a,b \in R$, what values of $n$ (which will probably be relating to $a$ and $b$) will make this inequality true?
$a^n+b>n^2$
I HAD a theory that it's related to $n=max$ {${a+|b|,\frac{|b|}{a-1}}$} works but I'm having trouble proving it.
If $a = 1 + t$, $$(1+t)^n \ge 1 + n t + {n \choose 2} t^2 + {n \choose 3} t^3 $$
so it is sufficient to have $n > -b/t$ and ${n \choose 3}/n^2 > 1/t^3$. Note that
$$ {n \choose 3}/n^2 = \frac{n}{6} - \frac{1}{2}+\frac{1}{3n}$$ so it is enough that $$n > \max\left(\frac{-b}{t}, 3 + \frac{6}{t^3} \right)$$.