2 to the n is Greater than n Cubed when n is 10 and above

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Theorem

$\forall n \in \Z, n \ge 10: 2^n > n^3$


Proof

Proof by induction:

For all $n \in \Z$ such that $n \ge 10$, let $P \left({n}\right)$ be the proposition:

$2^n > n^3$


We note that:

$2^9 = 512 < 729 = 9^3$

so when $n < 10$ the proposition does not hold.


Basis for the Induction

$P \left({10}\right)$ is the case:

$2^{10} = 1024 > 1000 = 10^3$

so $P \left({10}\right)$ is seen to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 10$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is our induction hypothesis:

$2^k > k^3$


We need to show that:

$2^{k + 1} > \left({k + 1}\right)^3$


Induction Step

This is our induction step:

We note that when $k \ge 10$:

\(\ds \left({1 + \dfrac 1 k}\right)^3\) \(=\) \(\ds 1 + \frac 3 k + \frac 3 {k^2} + \frac 1 {k^3}\) Binomial Theorem
\(\ds \) \(\le\) \(\ds 1 + \frac 3 {10} + \frac 3 {100} + \frac 1 {1000}\)
\(\ds \) \(=\) \(\ds 1 + \frac {300 + 30 + 1} {1000}\)
\(\text {(1)}: \quad\) \(\ds \) \(<\) \(\ds 2\)


Thus:

\(\ds 2^{k + 1}\) \(=\) \(\ds 2 \times 2^k\)
\(\ds \) \(>\) \(\ds \left({1 + \dfrac 1 k}\right)^3 2^k\) from $(1)$
\(\ds \) \(>\) \(\ds \left({1 + \dfrac 1 k}\right)^3 k^3\) Induction Hypothesis
\(\ds \) \(=\) \(\ds k^3 + \frac {3 k^3} k + \frac {3 k^3} {k^2} + \frac {k^3} {k^3}\)
\(\ds \) \(=\) \(\ds k^3 + 3 k^2 + 3 k + 1\)
\(\ds \) \(=\) \(\ds \left({k + 1}\right)^3\)

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Sources

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