2

I have the math problem (induction proof - $n!^2 > n^n$) that I try to solve and I haven't yet managed to get it right so maybe somebody could help me.

My current plan solving the problem is the following :

1) for : $n = 1 \implies$ False; $n = 2 \implies$ false; $n = 3 \implies$ true

2) for $n = k \implies (k!)^2 > k^k$

3) $1^2 \cdot 2^2 \cdot 3^2 \cdot \ldots \cdot (k*k) > k^k $

4) $1^2 \cdot 2^2 \cdot 3^2 \cdot \ldots \cdot k > k^{k-1}$ move one k to right part of inequality

5) $k > \frac{k^{k-1}}{[(k-1)!]^2}$

I am aware that this proof is not complete!
I would appreciate very much some help.

Thank you

CameronJWhitehead
  • 1,942
Mihai
  • 123

1 Answers1

2

We need to show: $k^k\cdot (k+1)^2 > (k+1)^{k+1}$ for $k > 3$.But it is the same as proving:

$\dfrac{(k+1)^2 }{k} > \left(\dfrac{k+1}{k}\right)^{k+1} \iff k+1 > \left(1+\dfrac{1}{k}\right)^{k}$. But this is true because: $k + 1 > 3 > \left(1+\dfrac{1}{k}\right)^k$, $\forall k > 3$

DeepSea
  • 77,651