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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, otherwise it is unclear what you mean. – ybungalobill Dec 25 '12 at 19:50

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... – DonAntonio Dec 25 '12 at 20:19
@DonAntonio: I did the opposite order so that it would be more natural to conclude the thing blows up. – André Nicolas Dec 25 '12 at 20:22
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. – DonAntonio Dec 25 '12 at 20:27
@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 $a_{N+m}\gt a_N(1.1)^m$. – André Nicolas Dec 25 '12 at 20:32
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. – DonAntonio Dec 25 '12 at 20:41

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 – DonAntonio Dec 25 '12 at 20:16
No, it's not since the the limit of the quotient, and thus also the lim sup, is greater than 1. – DonAntonio Dec 25 '12 at 20:20
oh yeah, sorry, i am stuck in thought, absolutely, you are right – doniyor Dec 25 '12 at 20:21

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