Finding the limit of a function with factorials

Question

I am trying to take the limit of this sequence and to do so, I must first rewrite the expression to be able to cancel out terms.

For example:

$\require{cancel}$ \begin{align}\lim_{n\to \infty}a_n&=\lim_{n\to\infty}\left(\frac{(2n-1)!}{(2n+1)!}\right)\\ &=\lim_{n\to\infty}\left(\frac{\cancel{(2n-1)}\cancel{(2n-2)}\cancel{(2n-3)}\cdots\cancel 3\cdot\cancel2\cdot\cancel1}{(2n+1)(2n)(\cancel{(2n-1)}\cancel{(2n-2)}\cancel{(2n-3)}\cdots\cancel 3\cdot\cancel2\cdot\cancel1}\right)\end{align}

What I understand:

$$(2n-1)! = (2n-1)(2n-2)(2n-3)\cdots$$

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What I don't understand:

Where does the $3\cdot2\cdot1$ come from?

Shouldn't the next 3 terms be $(2n-4)(2n-5)(2n-6)$?

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My reasoning:

\begin{align}(2n-3) - 1 &= (2n-4)\\ (2n-4) - 1 &= (2n-5)\\ (2n-5) - 1 &= (2n-6)\end{align}

Any help is greatly appreciated.

Answer 1

You need to bear in mind that a factorial eventually has to end. When the author places

$$(2n-1)! = (2n-1)(2n-2)(2n-3)\cdots(3)(2)(1)$$

those "$\cdots$" means "the obvious pattern continues until" ... whatever appears after the dots. The obvious pattern is that of the factorial: $z!$ is the product of all of the positive integers less than or equal to $z$ (if $z$ is itself a positive integer). We could write that as, for instance,

$$z! = z(z-1)(z-2)\cdots(3)(2)(1)$$

This becomes more obvious if you plug in a specific value for $n$ or $z$. Let $n=6$ for instance ($z=11$). Then

$$(2n-1)!=z!=11!=11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$$

though this is cumbersome to write, so one might instead use

$$11! = 11 \cdot 10 \cdot 9 \cdot \cdots \cdot 3 \cdot 2 \cdot 1$$

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This is to be contrasted with the case where you have nothing after the dots, in which "$\cdots$" means "the obvious pattern continues infinitely". For example,

$$1 + \frac 1 2 + \frac 1 4 + \frac 1 8 + \frac{1}{16} + \frac{1}{32} + \cdots$$

would be such an "infinite" case. (It is the summation $\sum_{n=0}^\infty (1/2)^n$.) Factorials however have to end.

Think of it like this: if the factorial did not end, you would eventually multiply by $0$, which would mean all factorials would be $0$. Kind of silly, isn't it? That's why the interpretation of what "$\cdots$" represents is important.

Answer 2

The correct way do do math is to never rely on ellipsis (aka. three dots). At best, they lead to ambiguity, e.g., how should the sequence $2,3,5,7,11,\ldots$ be continued? If you think the next term is $13$, it needn't be so. So intead, we should define factorial for all $n\in\Bbb N_0$ by a nice and unambiguous recursion: $$ n!:=\begin{cases}1&\text{if }n=0\\n\cdot(n-1)!&\text{otherwise}\end{cases}$$ With this, $$\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\cdot(2n)!}=\frac{(2n-1)!}{(2n+1)\cdot(2n)\cdot(2n-1)!}=\frac1{(2n+1)\cdot (2n)} $$ for all integers $n\ge1$.