# What does ! mean in sequences?

I'm doing a sequences problem where I have to write the first five terms of a sequence. It looks normal, but there is an exclamation mark on the denominator:

$$a_n = \frac{1}{(n + 1)!}$$

&

$$a_n = \frac{(-1)^nn}{n! + 1}$$

What does the exclamation mark mean, and how do I go about this differently with the exclamation mark?

-
$n!$ is the factorial of $n$. (See the link.) –  David Mitra Apr 9 '12 at 0:01
That’s the factorial sign: $n!=n(n-1)(n-2)\ldots\cdot2\cdot1$, and we define $0!=1$. Thus, $4!=4\cdot3\cdot2\cdot1=24$. –  Brian M. Scott Apr 9 '12 at 0:02
It means that sequences are very exciting! –  alex.jordan Apr 9 '12 at 1:06

The factorial of a number is represented by the exclamation point (!). The factorial of a number $x$ is often described as the product of all positive integers less then or equal to $x$. For example:

$$4! = 4\cdot3\cdot2\cdot1 = 24$$

It is often also useful to describe $x!$ in a recursive relation:

$$x! = x(x-1)!$$

where $0! = 1$.

This method is often good because it helps explain why $0!=1$ (see also the "empty product" for more on this). Using the above explanation, we may find $4!$:

$$4! = 4(3)! = 4(3)(2)! = \cdots = 4(3)(2)(1)(0)! = 4(3)(2)(1)(1) = 24$$

-
Awesome, thanks! –  Zolani13 Apr 9 '12 at 0:26

If the number is 4! then you count down so it would be $4\times 3 \times 2\times 1= 24$. Another thing is that if the number is 7! you would count down until you get to two. So it would be $7\times 6\times 5\times 3\times 2 =5040$

-
It might be confusing to someone who has never seen factorials before if you count down to 1 in one example and 2 in another example (even though the result is the same). –  Daenerys Naharis May 19 '14 at 23:27
@Hockeylover It's actually defined recursively and one "counts down" to $0! :=1$. –  Bye_World Jan 7 at 20:55