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In my textbook, in order to solve for the limit of the sequence $$\lim_{x \to \infty} \frac{n!}{n^n} \tag{1}$$

the book rewrites the sequence $$a_n=\frac{1}{n}\left(\frac{2\cdot3\cdot\,\cdots\,\cdot n}{n\cdot n\cdot\,\cdots\,\cdot n}\right)\tag{2}$$

But how can these two expressions be equal? If you plug in numbers $1$ through $4$ into the second equation, it doesn't make sense. For example, for $n=4$, $$\left(\frac{2\cdot3\cdot\,\cdots\,\cdot 4}{4\cdot 4\cdot\,\cdots\,\cdot 4}\right) \tag{3}$$ makes no sense: $n$ must be $\geq5$ because in the numerator (of the parentheses part) $$2\cdot3\cdot\,\cdots\,\cdot n \tag{4}$$ there are two multiplication dots separated by an ellipsis, implying that there is at least one value between $3$ and $n$. Thus, the second equation can only be used for $n$ must be $\geq5$.

My textbook says that $$\left(\frac{2\cdot3\cdot\,\cdots\,\cdot n}{n\cdot n\cdot\,\cdots\,\cdot n}\right) \tag{5}$$ can equal $1$ for $n=1$ but technically you have to be able to plug in $1$ for $n$ to say that.

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  • $\begingroup$ The conversion you write from $a_n=\frac {n!}{n^n}$ to $a_n=\frac{1}{n}\left(\frac{1\cdot2\cdot3\cdot...\cdot n}{n\cdot n\cdot n\cdot...\cdot n}\right)$ is off by a factor $\frac 1n$. I don't know what the leading $\frac 1n$ is doing in the second. $\endgroup$ – Ross Millikan Oct 19 '19 at 2:06
  • $\begingroup$ @RossMillikan My bad I mistyped. $\endgroup$ – user532874 Oct 19 '19 at 2:10
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    $\begingroup$ In a now-deleted answer, @fordjones makes the important point that the ellipsis notation represents a pattern and shouldn't be taken entirely literally. "$1\cdot 2\cdot 3\cdot\,\cdots\,\cdot n$" is a common way to express $n!$, even for $n=1$ through $n=4$, despite the notation suggesting that there are terms between "$3$" and "$n$". Without resorting to the somewhat scary-looking sigma notation, the ellipsis notation is simply the most-convenient way to express "the product of the integers from $1$ to $n$" in an arithmetical form. Mathematics, like any language, has its colloquialisms. $\endgroup$ – Blue Oct 19 '19 at 2:21
  • $\begingroup$ You could click the edit button at the bottom of the post and correct the error. $\endgroup$ – Ross Millikan Oct 19 '19 at 2:27
  • $\begingroup$ @RossMillikan I did $\endgroup$ – user532874 Oct 19 '19 at 2:28
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You are correct in that, strictly, what has been written makes no sense. That being said, these kinds of abuses of notation are common throughout mathematics. A more rigorous expression would be $$\frac{n!}{n^n}=\frac{\prod_{i=1}^n i}{\prod_{i=1}^n n}$$

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  • $\begingroup$ For $a_n=\frac{1}{n}\left(\frac{2\cdot3\cdot...\cdot n}{n\cdot n\cdot...\cdot n}\right)$ I'm still not sure what the domain is. It's a sequence so $n$ must be a positive integer but which positive integers are allowed, strictly speaking? $\endgroup$ – user532874 Oct 19 '19 at 2:07
  • $\begingroup$ Yes, $n$ should be a positive integer. This is another common abuse, some letters tend to mostly be used to denote objects of a certain kind. $\endgroup$ – Reveillark Oct 19 '19 at 2:20
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Are you sure about $$\lim_{x \to 2} \frac{n!}{n^n} ?$$

$$a_n=\frac{1}{n}\left(\frac{1\cdot2\cdot3\cdot...\cdot n}{n\cdot n\cdot n\cdot...\cdot n}\right) = \frac{n!}{n^n}$$

if the number of $n$ in the denominator of $$(\frac{1\cdot2\cdot3\cdot...\cdot n}{n\cdot n\cdot n\cdot...\cdot n})$$ is $n-1$ which does not seem so.

Otherwise it is not correct.

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  • $\begingroup$ I made a mistake typing. $\endgroup$ – user532874 Oct 19 '19 at 2:14
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It seems that in the book they are using that

$$a_n=\frac{1}{n}\left(\frac{2\cdot3\cdot\,\cdots\,\cdot n}{n\cdot n\cdot\,\cdots\,\cdot n}\right)=\frac1n\frac{n!}{n^{n-1}}=\frac{n!}{n^n}$$

and indeed for $n=1$ we have $\frac{n!}{n^{n-1}}=\frac{1}{1^0}=1$.

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