can someone help me with this limit?
I don't know how to expand that factorial multiplication, so what I've done so far is substitute what given:
$$ \lim_{n\to\infty} \frac{n!\times(2n)!}{(3n)!}$$
$$ \lim_{n\to\infty} \frac{\infty\times\infty}{\infty}$$
And with this I can apply the Cauchy or L'Hôpital's theorem by deriving both sides of the fraction independently, but my problem also stars here because, I don't how to derive a factorial term.
Can someone help me please?
Thanks
Best Answer
In this case I'd just write it out recursively. Denote by $a_n$ the $n^{th}$ term of the sequence, so that $$ a_n = \frac{n!(2n)!}{(3n)!} = \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)}a_{n-1}$$ The basic idea behind understanding sequences is finding something simpler to compare it to. So let's take apart that factor: $$ \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)} = \frac{1}{3}\frac{2n-1}{3n-1}\frac{2n-1}{3n-2} < \frac{1}{3} $$ since each of those latter two fractions is less than $1$.
That means every new term in the sequence is less than a third the previous term: $a_n < a_{n-1}/3$. Your sequence starts with $1$, so $a_n < (1/3)^n$.
So now, ask a simpler question: what does $(1/3)^n$ do? Then use the answer to answer your original question.