Induction: prove $2n^2 < 10\cdot n!$

Question

Prove that $2n^2 < 10\cdot n!$, where $n$ is a positive integer

My approach:

$P(1)$ is true, and I'm trying to prove that $2(k+1)^2 < 10 (k+1)!$

Assume $2k^2 < 10\cdot k!$,

and $2k^2 * (k+1) < 10 (k+1)!$.. then what should I do?

Answer 1

Then you need to prove that: $2(k+1)^2 < 2k^2\cdot (k+1)$ and this amounts to: $k+1 < k^2$ which is true since $k+1 < 2k < k^2$ for $2 < k$

Answer 2

Hint: $2n^2 < 10 (n!)$ is equivalent to $n<5(n-1)!$, and this is simpler to prove by induction.

Answer 3

Notice that for $n>4$ , you have that $(n-1)^2 >n , (n-2)^2 >n $ . Just find the solutions to $n^2 -(n-k)^2 <0$ , and you see that the solution set is $n>k/2$ ( you can use the greatest integer larger than $k/2$ ). Since , for $n>4$ we have $(n-1)^2 >n $ and $(n-2)^2 >n $ , we get (in this range): $$n^2 =(n)(n) < n[(n-1)(n-2)]< n (n^{1/2})n^{1/2} < n!=n[(n-1)(n-2)...2.1 $$ , i.e., that for $n>4$ , $$ n^2 < n!$$ . Now, multiply the left by 2, and the right hand side by $10$ , and the equality is preserved, and you get $$ 2n^2 <10n! $$, which is what was to be proved. The first 4 cases can be computed directly and then this argument can be used.