Let's say that I wanted to prove the following problem which is similar to a homework problem. That is, the numerator and denominator are different, but similar enough that you posting an answer would likely give everything away.

$$\lim_{n\to\infty} \frac{6n^4}{9\left \lfloor{\ln n}\right \rfloor!} = 0 $$

Using L'Hopitals rule does not help here (well, maybe), and factoring does not help either.

I am not sure where I could start. I want to solve this myself so I can solve my homework problem.

Is there anything obvious that jumps off the bat about this problem that I should be seeing? I come across stuff similar to this all of the time but never know where to start and I feel like I just get lucky after enough guesses, so I want to learn how to approach the problem.


If it helps, I actually do not care that the limit is 0. I just care about which one grows faster as $n$ gets large and being able to prove that.

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    $\begingroup$ You understand that the factorial in the denominator in general is of an irrational number. $\endgroup$ – Shailesh Sep 17 '15 at 2:23
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    $\begingroup$ @Shailesh, isn't it the factorial of the floor function? $\endgroup$ – Yeldarbskich Sep 17 '15 at 2:27
  • $\begingroup$ @yeldarbskich Sorry. I did not see the floor function $\endgroup$ – Shailesh Sep 17 '15 at 2:37

Consider what happens when you go from $n$ to $en$ in the numerator and denominator.

  • $\begingroup$ Perhaps you meant $e^n$ ? $\endgroup$ – Lucian Sep 17 '15 at 4:27
  • $\begingroup$ No. $\ln(ne) = \ln(n)+\ln(e) = \ln(n)+1$, so this increases what is in the $\lfloor \rfloor$ by one, so the next factorial will be computed. $\endgroup$ – marty cohen Sep 17 '15 at 5:41

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