Does the notation "$\frac{1}{n}\left(\frac{2\cdot3\cdot\,\cdots\,\cdot n}{n\cdot n\cdot\,\cdots\,\cdot n}\right)$" make sense for $n<5$?

Question

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.

Answer 1

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 $$\frac{n!}{n^n}=\frac{\prod_{i=1}^n i}{\prod_{i=1}^n n}$$

Answer 2

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.

Answer 3

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 $\frac{n!}{n^{n-1}}=\frac{1}{1^0}=1$.