How can we prove that $x_n$ tends to infinity as $n$ tends to infinity? I have tried to rewrite the expression for $x_n$ as $x_n = (2^nn!)^2/(2n)! =(2^n)^2n!/(n+1)(n+2)...(2n)>(2^n)^2n!/2^nn^n=2^nn!/n^n$ but can't go any further.
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$$ \frac{(2n)!!}{(2n-1)!!}=\prod_{i=1}^n\left(1+\frac{1}{2i-1}\right)\ge \prod_{i=1}^n\left(1+\frac{1}{2i}\right) \ge \frac{1}{2}\sum_{i=1}^n \frac{1}{i} \to \infty. $$
Paolo Leonetti
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Consider the ratio of consecutive terms:
$$\frac{X_{n+1}}{X_n}=\frac{\frac{(2n+2)!!}{(2n+1)!!}}{\frac{(2n)!!}{(2n-1)!!}}=\frac{(2n+2)!!(2n-1)!!}{(2n+1)!!(2n)!!}=\frac{(2n+2)!}{2n+1}\cdot\frac1{(2n+1)!}=\frac{2n+2}{2n+1}$$
Thus,
$$X_{n+1}=X_1\cdot\prod_{k=1}^n\frac{X_{k+1}}{X_k}=2\prod_{k=1}^n\left(1+\frac1{2k+1}\right)\ge 2\sum_{k=1}^n\frac1{2k+1}\;,$$
and $\sum_{k=1}^\infty\frac1{2k+1}$ of course diverges.
Brian M. Scott
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Brian, isn't $(2n+2)!!/(2n)!!=2n+2$? I think the OP is referring to the semifactorial https://en.wikipedia.org/wiki/Double_factorial – Paolo Leonetti Oct 18 '16 at 12:37
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@Paolo: Of course it’s the double factorial: I merely misread a sign in the numerator and did a simplification that didn’t exist. Once that’s corrected, it becomes very similar to your result, so I may end up deleting it. – Brian M. Scott Oct 18 '16 at 12:50
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You can just add the remaining part, that's fine ;) – Paolo Leonetti Oct 18 '16 at 12:52