vanhees71 said:
Just write out the factorials. Then you'll see immediately how to simplify the expression.
I cannot tell from writing them out.
$$\frac{1*3*5*7*9...(2n-1)!}{3*5*7*9*11...(2n+1)!} $$
Please explain.Edit:
Nvm, I understand. Thank you!
$$\frac{(2n-1)!}{(2n+1)!} = \frac{(2n-1)*(2n-2)*(2n-3)*(2n-4)...2*1}{(2n+1)*(2n)*(2n-1)*(2n-2)...2*1} $$
$$ = \frac{1}{(2n)*(2n+1)} $$
or
$$\frac{(2n-1)!}{(2n+1)!} = \frac{(2n-1)!}{(2n+1)*(2n)*(2n-1)!} $$
$$ = \frac{1}{(2n+1)*(2n)} $$
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