4
$\begingroup$

Show that if $n ≥ 6$, then $n! > n^3$

Initial Step: $n = 6$

LHS: $6!=720$ RHS: $6^3=216$ LHS > RHS

Inductive Step: Assume $n=k$ is true

$k! > k^3$

Prove $n=k+1$ is true

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

Can you help me? I don't know where to go from here. I'm stuck in here.

$\endgroup$
1
  • $\begingroup$ Can you prove $(7/6)^3<2\le k+1$ for $k\ge6$? What does that tell you about $((k+1)/k)^3$? $\endgroup$ – J.G. Oct 6 '20 at 16:53
2
$\begingroup$

Assume that $k! > k^3$ ...$(1)$

Let's show $(k+1)k! > (k+1)^3$

Multiply $(1)$ by $(k+1)$

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

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

Using

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

$k^3-k^2+2k-1>0$, $k^3>k^2$, and, $2k-1>0$ given that $k\geq6$

$\endgroup$
1
$\begingroup$

It suffices to show that $k^3>(k+1)^2$ for $k≥6$ and you can do that using induction.

$\endgroup$
0
$\begingroup$

Base case of $k=6$ proved.

Moving forward, LHS increases by a factor of $(k+1).$

RHS increases by a factor of $\left(\frac{k+1}{k}\right)^3.$

Although it can be argued that $k=7$ and $k=8$ must be independently verified, thereafter, with $\frac{k+1}{k} < 2$,

RHS increasing by smaller factor than LHS.

$\endgroup$
0
$\begingroup$

\begin{align} (k+1!) &> (k+1)^3 \\ k! (k+1) &> (k+1)^3 \\ k! &> (k+1)^2 \end{align}

Our inductive hypotesis is $k! > k^3$.

$$ k^3 > (k+1)^2 $$ $$ k^3 - (k+1)^2 > 0 $$ (1)

$2.148$ is a root of $ f(k) = k^3-(k+1)^2$. Dividing $f(k)$ by $(x - 2.148)$ you can see that the discriminant of the second order polinomial is negative, so it's the only root. Also, it's derivarive $f'(k) = 3k^2 - 2(k+1)$ is positive for $k> 1.215$.

So, (1) is true for $k> 2.148$. In particular, for $k>6$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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