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I want to conclude about $\sup A$ and $\inf A$ where $A = \{ {n^n \over (n!)^2}: n = 1, 2, \dots \}$. My intuition is that $\sup A = 1$ and $\inf A=0$. To show that the infimum exists, I could prove that ${n^n \over (n!)^2}$ can be "as small as we wish" and compare it with $\inf A + \epsilon$ for whatever $\epsilon$ we choose. The problem here is that limits haven't been properly introduced in the material I'm going over. How can I go about this?

celtschk
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Zelazny
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  • You don't need a limit. You just have to show that no number $>0$ is a lower bound. Since $0$ is obviously a lower bound, and it is quite obviously the largest number not greater than $0$, it then follows that $0$ is the infimum. – celtschk Oct 27 '16 at 19:21
  • This basically reiterates what I wrote. I am interested in a more formal approach. One I considered was induction to show $(n!)^2} \ge n^n$ but it wouldn't be enough to conclude that this indeed can be as small as we choose. – Zelazny Oct 27 '16 at 19:29
  • My point is that there is no limit involved. – celtschk Oct 27 '16 at 19:30
  • "The problem here is that limits haven't been properly introduced" this is only a device by which I say "I know it's fairly obvious this goes to 0. Limits would give me the desired answer. I cannot think of any other formal technique of the top of my head." – Zelazny Oct 27 '16 at 19:37
  • No. That the limit is $0$ (or that a limit even exists) is neither necessary nor sufficient for proving the infimum. I can give you a sequence that doesn't converge to $0$ but has $0$ as infimum, and another sequence that converges to $0$ but does not have $0$ as infimum. – celtschk Oct 27 '16 at 19:45
  • I know that limits are not necessary. I wrote that. However, if I know that the expression comes very close to $0$ for sufficiently large numbers, this simplifies the use of the definition. In this particular case, I would need to show that for all $\epsilon >0$ there exists $x \in A$ such that $inf \text{A} + \epsilon > x$. This simplifies to $x < \epsilon$, which tells me that if I can choose $n$ for which the inequality is satisfied, I'm done. The limit is not necessary, but in this case is quite helpful and I'm looking for a way to show that $x$ comes very close to $0$. – Zelazny Oct 27 '16 at 19:54
  • But that is not a limit. – celtschk Oct 27 '16 at 19:56
  • $\lim \limits_{n \to \infty} {n^n \over (n!)^2} = 0$ gives me $x$ very close to $0$ for sufficiently large $n$, closer than $\epsilon$ is. – Zelazny Oct 27 '16 at 20:04
  • So does $\lim \limits_{n \to \infty} {(-n)^n \over (n!)^2} = 0$. But the infimum of ${{(-n)^n \over (n!)^2}:n=1,2,\ldots}$ is definitely not $0$. On the other hand, $a_n={n^n \over (n!)^2}+1+(-1)^n$ doesn't converge at all, but ${{n^n \over (n!)^2}+1+(-1)^n:n=1,2,\ldots}$ has infimum $0$. – celtschk Oct 27 '16 at 20:23

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