my textbook shows the result of n!/(n!+1) to be $\displaystyle \frac{1}{n+1}$

Does anyone know how they arrive at this result? Even wolfram shows something different.

also latex is throwing errors whenever i use factorials.

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- Apr 21st 2011, 08:11 AMskyd171factorial long division (or not?)
my textbook shows the result of n!/(n!+1) to be $\displaystyle \frac{1}{n+1}$

Does anyone know how they arrive at this result? Even wolfram shows something different.

also latex is throwing errors whenever i use factorials. - Apr 21st 2011, 08:13 AMProve It
It's actually n!/(n + 1)! = 1/(n + 1). This is because (n + 1)! = (n + 1)n!.

- Apr 21st 2011, 08:21 AMskyd171
Thanks. why is it equal to that? is this a rule for distributing factorials over parentheses? i have not seen that rule before.

- Apr 21st 2011, 08:22 AMProve It
It's because N! = N(N - 1)(N - 2)...(3)(2)(1) = N(N - 1)!.

What happens if N = n+1? - Apr 21st 2011, 08:28 AMskyd171
should be N!= n+1(n)(n-1)(n-2)...(4)(3)(2)(1)=N+1(N)!

- Apr 21st 2011, 08:32 AMProve It
First of all, you MUST use brackets where they are necessary and second, N and n are different.

It should be N! = (n + 1)(n)(n - 1)...(3)(2)(1) = (n + 1)n!.

Therefore (n + 1)! = (n + 1)n! - Apr 21st 2011, 08:32 AMPlato
What you posted, http://quicklatex.com/cache3/ql_7484...4e9e03f_l3.png, is not true if $\displaystyle n\ne 1$.

It is true that

http://quicklatex.com/cache3/ql_b9ad...00d260f_l3.png - Apr 21st 2011, 09:13 AMskyd171
- Apr 21st 2011, 09:31 AMPlato
I am not sure how to read that.

But the is true:

http://quicklatex.com/cache3/ql_07fc...c33781e_l3.png