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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.

Riley H
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  • Try $a=1.5, b=0$ – gammatester Oct 08 '18 at 16:19
  • Shoot you're right, that disproves my idea. Do you have any clue how to solve this algebraically? – Riley H Oct 08 '18 at 16:29
  • What is $n$? An integer? A positive integer? Any real number? Note: If $b>-1$ and $n=0$ the equation is also satisfied. – Martin Rosenau Oct 08 '18 at 16:37
  • IMO you have to solve it numerically. For large $n$ you know than $a^n$ grows – gammatester Oct 08 '18 at 16:37
  • If this is a rehashing of your previous question note that in your original question we don't care at all about finding the smallest value of $n$ making that true, we only cared about finding at least one $n$ for which it is true (and is true for all other values of $n$ above it), or more accurately, proving that such an $n$ exists (and not even bother explicitly finding it). – JMoravitz Oct 08 '18 at 16:39

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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 $$n > \max\left(\frac{-b}{t}, 3 + \frac{6}{t^3} \right)$$.

Robert Israel
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