Prove by induction for $n \geq n_0$, $n^3 < 3^n$. What is the value of $n_0$?
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1http://math.stackexchange.com/questions/434505/show-by-induction-that-n3-leq-3n-for-all-natural-numbers-n – lab bhattacharjee Nov 08 '13 at 07:57
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for n=3, n^3=3^n. Does this help ? – Claude Leibovici Nov 08 '13 at 07:58
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@KennedyS It also may be helpful for you to look at$. It will help you in formatting. – Jeremy Upsal Nov 08 '13 at 08:00
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$n^3 < 3^n$ when $n \ge 4$, so $n_0 = 4$. This is our base case since $64 < 81$.
Assume the result to be true for $n=k$, then $k^3 < 3^k\implies 3k^3 < 3^{k+1}$.
We want to show $(k+1)^3 = k^3 + 3k^2 + 3k + 1 \le 3k^3$ for $k\ge4$.
So just show that $2k^3 - 3k^2 - 3k - 1 \ge 0$ for $k\ge4$.
