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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

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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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    In fact I think that looking at the inverse $$\frac{a_n}{a_{n+1}}$$ can be more helpful...2012-12-25
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    @DonAntonio: I did the opposite order so that it would be more natural to conclude the thing blows up.2012-12-25
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    I can't see how. The OP, just as in doniyor's answer, with get that the limit of that quotient is greater than 1 so that the series doesn't converge...but how from this *directky* is possible to deduce *the sequence* blows up? With the inverse quotient we get the series converges thus the general term's sequence converges to zero *and then* the original sequence, being a positive one, blows up to $\,\infty\,$...unless you had something else in mind.2012-12-25
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    @DonAntonio: I am not thinking **at all** in terms of the associated series. Let $N$ be so large that the ratio is say $\gt 1.1$. Then $a_{N+m}\gt a_N(1.1)^m$.2012-12-25
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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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    Thus the series $$\sum_{n=1}^\infty\frac{(2n-1)!}{3^n(n!)^2}$$doesn't converge...so??2012-12-25
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    No, it's not since the the limit of the quotient, and thus *also* the lim sup, is greater than 1.2012-12-25
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    oh yeah, sorry, i am stuck in thought, absolutely, you are right2012-12-25