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Let us take the most basic example:

$\lim_{n\to\infty} \frac{n!}{n^n}$

$\frac{n!}{n^n} = \frac{n(n-1)...1}{n \cdot n...n} = \frac{n}{n} \cdot \frac{n-1}{n} \cdot [...] \cdot \frac{1}{n} \leq \frac{1}{n}$ We used the fact that each term was smaller or equal than 1, so the result is upper-bounded by $\frac{1}{n}$, and its limit goes to 0.

This is all fine and good; the question is whether we are allowed to use this trick within a limit. In the above example, we've shown the inequality for $n \in \mathbb{N}$, and then we took its limit. Am I right in saying that it would be wrong to write: $\lim_{n\to\infty}\frac{n!}{n^n} = \lim_{n\to\infty}\frac{n(n-1)...1}{n \cdot n...n} = \lim_{n\to\infty}\frac{n}{n} \cdot \frac{n-1}{n} \cdot [...] \cdot \frac{1}{n} \leq \lim_{n\to\infty}\frac{1}{n}$ ?

In this second case, there are actually an infinite amount of terms $\frac{i}{n}$ for $ i \in \mathbb{N}$, therefore when need to take the limit of each term (infinitely many). Intuitively, it's clear why this would still 'work' in this example because each term has a limit, but to me it seems that this isn't formal... In the same sense that we cannot deal with infinite dimensional vector spaces the way we can with finite ones, is there a similar difference between expressions containing uncountably infinitely many terms and those containing countably many?

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    $\begingroup$ It would be better to avoid limit signs altogether. You've shown $0\le n!/n^n \le 1/n$ for all $n$. By the Squeeze Thm, you're done. $\endgroup$ – zhw. Nov 3 '15 at 21:31
  • $\begingroup$ @zhw : Thank you for your answer. Yes, I agree. But this is not what this question is about. It is about the formalism of limits with infinitely many terms. $\endgroup$ – Symeof Nov 3 '15 at 21:37
  • $\begingroup$ There are never infinitely many terms $\frac{i}{n}$ because $n$ is always finite. It is a fact that if $a_n \le b_n$ for all $n$ and both are convergent sequences then $\lim a_n \le \lim b_n$. So your reasoning is justified. This statement is essentially the meat of the squeeze theorem. $\endgroup$ – Reinstate Monica Nov 3 '15 at 21:43
  • $\begingroup$ @Solomonoff's Secret : So you are saying that the number of terms in the limit of $n^n$ is finite? $\endgroup$ – Symeof Nov 4 '15 at 10:34
  • $\begingroup$ I think you are confused about something. $\lim_{n\to\infty} \frac{n}{n} \cdot \ldots \cdot \frac{1}{n}$ does not mean the limit of the construction $\frac{n}{n} \cdot \ldots \cdot \frac{1}{n}$, then numerically evaluated. It means the limit of a sequence of numbers, each number equaling an expression of the form $\frac{n}{n} \cdot \ldots \cdot \frac{1}{n}$. Each of these expressions has finitely many terms. Each expression evaluates to a number. And the limit of these numbers is a number. There are no infinite expressions here. $\endgroup$ – Reinstate Monica Nov 4 '15 at 13:11

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