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$\lim_{n\to\infty}\frac{(2n-1)!}{3^n(n!)^2}$

How can I associate limit problem with series? And how can i find limits from series? Can anyone help?

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    Please write the formula using [LaTeX](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117), otherwise it is unclear what you mean.2012-12-25

2 Answers 2

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Hint: Let $a_n=\dfrac{(2n-1)!}{3^n(n!)^2}$.

It is useful to look at the ratio $\dfrac{a_{n+1}}{a_n}$ for large $n$.

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    Well, but then you're not aiming at "associate the limit with series", which is what the OP asked and what I thought you were trying to do. Ok, thanks for the clarification.2012-12-25
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by ratio rule:
$\dfrac{(2(n+1)-1)!}{3^{n+1}((n+1)!)^2}\cdot\dfrac{3^n(n!)^2}{(2n-1)!}= \dfrac{4n^2+2n}{3n^2+6n+3} \rightarrow \dfrac{4}{3}$

thus the series doesnot converge as the quotient and thus limsup is bigger than 1

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    oh yeah, sorry, i am stuck in thought, absolutely, you are right2012-12-25