<s>[INST] For what natural numbers is $n^3 < 2^n$? Prove by induction Problem
For what natural numbers is $n^3 < 2^n$?
Attempt @ Solution


*

*For $n=1$, $1 < 2$

*Suppose $n^3 < 2^n$ for some $n = k \ge 1$

*It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$

*But, how can I prove this through induction?

 [/INST] You need to gather more data first:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9&10&11\\
n^3:&1&8&27&64&125&216&343&512&729&1000&1331\\
2^n:&2&4&8&16&32&64&128&256&512&1024&2048
\end{array}$$
Notice that $n^3$ is not less than $2^n$ for $n=2,3,\ldots,9$; the fact that $1^3<2^1$ is an isolated success. The simplest reasonable guess at this point is that $n^3<2^n$ if $n\ge 10$. Thus, the basis step for your induction will be the calculation that $10^3<2^{10}$.
For your induction step you’ll assume as induction hypothesis that $k^3<2^k$ for some integer $k\ge 10$. Your goal, then, will be to show that $(k+1)^3<2^{k+1}$. Note that
$$(k+1)^3=k^3\cdot\frac{(k+1)^3}{k^3}=k^3\left(\frac{k+1}k\right)^3\;.$$
Since $k\ge 10$, $\frac{k+1}k=1+\frac1k\le 1+\frac1{10}=\frac{11}{10}$, and therefore 
$$\left(\frac{k+1}k\right)^3\le\left(\frac{11}{10}\right)^3=\frac{1331}{1000}<2\;,$$
and therefore $(k+1)^3<2k^3$. By the induction hypothesis $k^3<2^k$, so ... ?
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