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I have $$s_1 = 1, s_n = ns_{n-1}$$

I don't know what this means at all, sequence 1 equals 1, sequence number = number times sequence subscript number - 1

Is that it? Because it doesn't work at all when I try to work it out.

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I am glad to see you posting questions, this will be more beneficial for you. – picakhu Apr 9 '11 at 16:45

We use $s_n=ns_{n-1}$: $$s_1=1$$ $$s_2=2s_{2-1}=2s_1=2$$ $$s_3=3s_2=3\cdot 2=6$$ $$s_4=4s_3=4\cdot 6=4\cdot3\cdot 2=24$$

The idea is that we're given a starting number, and a procedure that will get us from one number to the next. We have a starting position and a way to move forward. This in total gives us our entire sequence. In this case it's relatively easy to see that we're multiplying all natural numbers up to $n$ (called $n!$ or "n factorial"), as joriki said. $s_n$ is not the sequence as a whole, it's the $n^{th}$ element of the sequence.

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Referring to $s_1$ as "sequence one" indicates your confusion. It is more accurately referred to as the "first element of the sequence." The sequence is the entire list of values $\{s_1,s_2,...,s_n,...\}$.

So we define the first element of the sequence, and then we define, for any $n$, the $n$th element of the sequence in terms of the $(n-1)$th term of the sequence.

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nitpick: the sequence is ordered, sets are not ordered. – Yuval Filmus Apr 9 '11 at 21:11
Yeah, I wasn't using the curly braces to indicate a set. – Thomas Andrews Apr 9 '11 at 21:33

I take it you mean $s_1=1, s_n=ns_{n-1}$. This is a recursive definition of the factorial, $s_n=n!$. The factorial of any natural number is that number times the factorial of its predecessor. For instance, $5!=5\cdot 4!$. Unwinding the recursion, you find that this is equivalent to saying that the factorial of a natural number is the product of all natural numbers up to and including that number. For instance,

$$ \begin{eqnarray} 5!&=&5\cdot4! \\ &=& 5\cdot(4\cdot3!)\\ &=& 5\cdot(4\cdot(3\cdot2!))\\ &=& 5\cdot(4\cdot(3\cdot(2\cdot 1!)))\\ &=& 5\cdot(4\cdot(3\cdot(2\cdot 1)))\\ &=& 5\cdot4\cdot3\cdot2\cdot1\;. \end{eqnarray} $$

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Joriki, you could edit the question to make it clearer. – Yuval Filmus Apr 9 '11 at 16:02
@Yuval: I considered that, but didn't know how to rewrite the second paragraph that uses "number" instead of a variable, and so thought it might be easiest to just clarify it in my own words in the answer. – joriki Apr 9 '11 at 16:05
Thanks I think I understand it now, it really just is not intuitive as to how these numbers are suppose to be used. – Adam Apr 9 '11 at 16:09

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