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How can I work with this? $$\frac{(3n)!}{(3(n+1))!}$$ I really don't know how to open this fatorial and then, simplify it.

Actually, I have to calculate the limit when $n\to\infty$.

Thanks :)

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1 Answer

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$$(3(n + 1))! = (3n + 3)! = (3n)!(3n + 3)(3n+2)(3n+1)$$

$$\frac{(3n)!}{(3(n+1))!} = \frac{\require{cancel}\cancel{(3n)!}}{\cancel{(3n)!}(3n + 3)(3n+2)(3n+1)} = \dfrac 1{(3n + 3)(3n+2)(3n+1)}$$ which goes to $0$ as $n\to \infty$.

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