Solving Limit Problem: n→∞, n!-1/n³ln(n!)

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SUMMARY

The limit of the expression \(\lim_{n\to\infty} \frac{n!-1}{n^{3} \ln(n!)}\) approaches infinity. This conclusion is derived from the inequality \(\ln(n!) \leq \frac{n(n + 1)}{2}\) and the established fact that \(\lim_{n \rightarrow \infty} \frac{n!}{n^{m}} = \infty\) for all integers \(m > 0\). The analysis shows that as \(n\) increases, the factorial function \(n!\) grows significantly faster than the polynomial term \(n^3 \ln(n!)\), confirming the limit's behavior.

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Lisa91
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Could anyone tell me please why the limit of this guy is infinity?

\lim_{n\to\infty} \frac{n!-1}{n^{3} \ln(n!)}
 
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Lisa91 said:
Could anyone tell me please why the limit of this guy is infinit
\lim_{n\to\infty} \frac{n!-1}{n^{3} \ln(n!)}
See what you can do with this inequality.

$\ln \left( {n!} \right) = \sum\limits_{k = 1}^n {\ln (k)} \leqslant \sum\limits_{k = 1}^n k =\frac{{n(n + 1)}}{2}$
 
The 'core' of the problem is to demonstrate that...

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!}{n^{m}} = \infty$ (1)

... for all integers m>0. That is easily achieved supposing n>m, writing ...

$\displaystyle \frac{n!}{n^{m}}= \frac{n\ (n-1)\ (n-2)\ ...\ (n-m+1)}{n^{m}}\ (n-m)\ (n-m-1)\ ...\ 2 = $

$\displaystyle = 1\ (1- \frac{1}{n})\ (1-\frac{2}{n})\ ... (1-\frac{m-1}{n})\ (n-m)\ (n-m-1)\ ...\ 2$ (1)

... and observing what happens if n tends to infinity...

Kind regards$\chi$ $\sigma$
 

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