Prove that $x_n = (2n)!!/(2n-1)!!$ sequence is divergent.

Question

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.

Answer 1

$$ \frac{(2n)!!}{(2n-1)!!}=\prod_{i=1}^n\left(1+\frac{1}{2i-1}\right)\ge \prod_{i=1}^n\left(1+\frac{1}{2i}\right) \ge \frac{1}{2}\sum_{i=1}^n \frac{1}{i} \to \infty. $$

Answer 2

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.