I am reading through a proof of $4^n > n^4$ for $n \geq 5$ using induction.

For the induction step, $4^{n + 1} > (n + 1)^4$ must be proven, and we know that $4^{n + 1} > 4n^4$ from the induction hypothesis. (All for $n \geq 5$, of course.)

The proof says that it is adequate to show that that $4n^4 \geq (n + 1)^4$ for $n = 5$, but how do we know that $4^{n + 1} > 4n^4 \geq (n + 1)^4$? Can't it be something like $4^{n + 1} > (n + 1)^4 > 4n^4$ since the only fact we know is that $4^{n + 1} > 4n^4$, and this is true for $4^{n + 1} > (n + 1)^4 > 4n^4$ also (or $4^{n + 1} > (n + 1)^4 \geq 4n^4$)?

And why is it $4n^4 \geq (n + 1)^4$ and not $4n^4 > (n + 1)^4$? (Greater than or equal to vs. strictly greater than)


To prove $4n^4 \geq (n+1)^4$, we can prove $$(1 + \frac{1}{n})^4 \leq 4$$ which is true as long as $n \geq 3$, since the LHS in strictly decreasing in $n$

And as you can see, the inequality can of course be made strict.

  • $\begingroup$ @nablablah Yes you can write it in the strict way, as I said at the end of my answer $\endgroup$ – Petite Etincelle Oct 6 '14 at 9:11

$4n^{4}-\left(n+1\right)^{4}=\left(2n^{2}-\left(n+1\right)^{2}\right)\left(2n^{2}+\left(n+1\right)^{2}\right)=\left(n^{2}-2n-1\right)\left(3n^{2}+2n+1\right)$ so it is enough to show that $n^{2}-2n-1\geq0$ for $n\geq5$.

Note that $n^{2}-2n-1=\left(n-1\right)^{2}-2$ showing that it is enough to show that $m^2\geq 2$ for $m\geq4$.

Sure you can take it from here :)


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