Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Not sure where to go with this

$2^k > k^3$ for $k > 9$

$2^(k+1) > (k+1)^3$

$2^(k+1) = 2^k \cdot 2$

$2^k \cdot 2 > k^3 \cdot 2$ (by inductive hypothesis)

$2^k \cdot 2 > 2k^3$

$2k^3 > k^3 + 3k^2 + 3k + 1$

$2k^3 > (k+1)^3$

I know the final inequality is true, since I graphed it, but I was wondering if there was a clearer way to show the thought process algebraically.

share|improve this question
add comment

2 Answers 2

You have $2^{k+1}=2\cdot2^k>2k^3$, so it suffices to show that $2k^3\ge(k+1)^3$, or, equivalently, that $k^3\ge 3k^2+3k+1$ for $k>9$. And

$$3k^2+3k+1<3\left(k^2+k+1\right)\le3\left(3k^2\right)=9k^2<k^3$$

for $k>9$.

share|improve this answer
add comment

You want to show that $2^k > k^3$ implies $2^{k+1} > (k+1)^3$ for $k > 9$.

Let's try proof by contradiction.

Suppose $2^k > k^3$ and $2^{k+1} \le (k+1)^3$. Then $(k+1)^3 \ge 2^{k+1} = 2\ 2^k > 2\ k^3 $ or $k^3 < 3k^2+3k+1$ which, by a non-coincidence, is the same as Brian M. Scott's inequality, and can be shown false the same way for $k > 9$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.