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I have the limit $$\lim \limits_{n \to \infty}\frac{n^3-n^2-2}{n!}.$$ I am able to convert it into the form $$\lim \limits_{n \to \infty}\frac{1}{\frac{n!}{n^3}}$$ but am stuck on how to solve $$\lim \limits_{n \to \infty}\frac{n!}{n^3}.$$ I know that $n!$ grows at a much larger rate than $n^3$ but am unsure how to express this in mathematical terms.

Guy Fsone
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Cosmic
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    $n! \ge n(n - 1)(n - 2)(n - 3) = n^4 + \text{ errors}$ once $n \ge 4$. –  Feb 14 '18 at 00:05
  • and what is $x$? has it some connection to $n$? – user Feb 14 '18 at 00:07
  • Sorry, formatted incorrectly. x is not a variable that is part of this question. – Cosmic Feb 14 '18 at 00:08
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    You may use that for large enough $n$ one has $n!>e^n$ (this is an incredibly loose comparison. for a tighter comparison, look up sterling's approximation) and so compare to the related limits $\frac{n^3-n^2-2}{e^n}$ and $\frac{e^n}{n^3}$. You should know that exponentials are a much faster growing variety than polynomials are. Also an option is to use L'Hopital's rule, (why I suggested comparison to $e^n$) – JMoravitz Feb 14 '18 at 00:22
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    Factorials grow faster than polynomials. And $\frac {1}{\frac {n!}{n^3}} = \frac {n^3}{n!}$ you have that inverted. – Doug M Feb 14 '18 at 00:26
  • To understand $\displaystyle\frac{n!}{n^3}$ rewrite it as $\displaystyle\left(\frac{n(n-1)(n-2)}{n^3}\right)(n-3)!$ and recognize that the first factor $\to 1$ and the second factor $\to \infty$ as $n\to \infty$. – Malcolm Feb 14 '18 at 01:27
  • @Cosmic Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 27 '18 at 16:27

4 Answers4

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Note that for $n$ large the expression is positive and $n!>n^4$ (you can prove easily a part this two assumptions) then

$$0\le\frac{n^3-n^2-2}{n!}\le\frac{n^3-n^2-2}{n^4}=\frac1{n}-\frac1{n^2}-\frac2{n^4}\to 0$$

thus for squeeze theorem

$$\lim \limits_{n \to \infty}\frac{n^3-n^2-2}{n!}=0$$

user
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  • https://math.stackexchange.com/q/2678602/456510 – MathLover Mar 06 '18 at 21:25
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    @MathLover Dear MathLover, Thanks for your kind words, I indeed saw your OP and I've noticed that it has received a lot of good aswers, in praticular from Jack and Paramanand Singh two of the most quoted and skilled people here on MSE. I will take alook and I'll come for a feedback to you. Bye – user Mar 06 '18 at 21:50
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Hint

$$\frac{n^3-n^2-2}{n!}=\frac{1}{(n-3)!}+\frac{2}{(n-2)!}-\frac{2}{n!}$$

Community
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E.H.E
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Another hint:

Note first that $\;u_n\sim_\infty\dfrac{n^3}{n!}$, so

$$\frac{u_{n+1}}{u_n}\sim_\infty \biggl(\frac{n+1}n\biggr)^3\frac1{n+1}\sim_\infty \frac1{n+1}\to 0.$$

Bernard
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As other answers have noted, there are plenty of fancy ways to show the limit you want to show. But remember, you can always go back to the definition of a limit and work from there. If we wish to show that

$$\lim \limits_{n \to \infty}\frac{n!}{n^3}$$

goes to infinity, we must show that for any choice of lower bound $L$ there exists an integer $m$ such that $m<n$ implies that $L<\frac{n!}{n^3}$

We begin by noting that $1 < \frac{(n)(n-1)(n-2)(n-3)}{n^3}$ for all $n>6$. (Can you prove this fact?)

Therefore, for our $m$ we can choose $L+4$ or $6$, whichever is larger, and we have

$$\frac{n!}{n^3} = (\frac{(n)(n-1)(n-2)(n-3)}{n^3})(n-4)(...) $$

if $n>m$ then we have the first part being greater than one, the $(n-4)$ part being greater than $L$, and the $(...)$ part being greater than one, so we can conclude that the whole product is greater than $L$, and we're done.

Going back to the definition is tedious but it gets the job done, and it's good practice.

Eric Lippert
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