If $n!>n^3$, find the value of $n$.

Question

I have a problem of getting $n$, I do not even know if it is possible. If $\space n! > n^3 \space so \space n! = \left( \frac { n ( n + 1 ) } {2}\right ) ^ 2$

I tried different methodologies and still have nothing to stick with. I do not know which topic I should dig in to solve this problem.

Could you solve it with an explanation ?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial, and equation editing how-to. — Jakeup, Oct 16 '21 at 06:58
To be clear, are you trying to find an example of some specific $n$ which satisfies the inequality after you wrote "if", and also the equation you wrote after "so"? [if so, what you have now doesn't convey that very well] — coffeemath, Oct 16 '21 at 07:12
Hence your problem is to solve $\left(\frac{n(n+1)}{2}\right)^2> n^3$? — Marcos, Oct 16 '21 at 07:16
In your comment you wrote $!n$ but in question it is $n!.$ Which is it? because $!n$ has a different meaning... — coffeemath, Oct 16 '21 at 07:17
i tried to proof that n! = (n(n+1)2)2 and it is not true in this case @MarcosEscartínFerrer — Cryptorrior, Oct 16 '21 at 07:19
@MarcosEscartínFerrer I think in addition OP wants the factorial of $n$ to be equal to the squared sum of first $n$ integers. [Don't know if that's powsible but I think OP is looking for that] — coffeemath, Oct 16 '21 at 07:19
@MarcosEscartínFerrer That simplifies to $(n-1)^2>0$, i.e. $n\ne1$. It's a long way from solving the original problem. — J.G., Oct 16 '21 at 07:20
I used wolfram alpha and $\left(\frac{n(n+1)}{2}\right)^2> n^3$ gave me $n\neq 1$ while $n!=\left(\frac{n(n+1)}{2}\right)^2$ gave me $n=1$, so the solution is that there are no positive integer I guess. — Marcos, Oct 16 '21 at 07:27
i am also trying different methodologies and got the same conclusion @MarcosEscartínFerrer — Cryptorrior, Oct 16 '21 at 07:36

Claude Leibovici · Answer 1 · 2021-10-16T12:46:16.223

The simplest would be to try with your pocket calculator.

Let us make the problem more general and say that we want to solve for $n$ the equation $$n!=a^n$$

If you have a look at this question of mine, you will see a superb approximation given bt @robjohn. Adapted to the problem ($k=0$), we have $$n \sim a\, e^{1+W(t)}-\frac 12\qquad \text{where} \qquad t=-\frac{\log (2 \pi a)}{2 e a}$$ where $W(t)$ is Lambert function.

Working with real numbers, for $a=3$, this gives $n=6.00778$ while the exact solution given by Newton method is $n=6.01602$.

Since $t$ is small for $a>1$, you can use $$W(t)=t-t^2+\frac{3 t^3}{2}-\frac{8 t^4}{3}+O\left(t^5\right)$$ $$e^{W(t)}=1+t-\frac{t^2}{2}+\frac{2 t^3}{3}-\frac{9 t^4}{8}+O\left(t^5\right)$$ or, even better $$e^{W(t)}=\frac{101 t^2+174t+60}{17 t^2+114t+60}$$

Using the above, for $a=3$, you would obtain $n=6.00868$.

If $n!>n^3$, find the value of $n$.

1 Answers1