I have the following expression $$ x_N = N\left( \frac{(2N-1)!!}{(2N)!!} \right)^2$$ In order to prove that the sequence is convergent one can observe that $\frac{x_N}{x_{N-1}} = \frac{(2N-1)^2}{4N(N-1)} = 1 + \frac{1}{4N^2 - 4N}$ hence $$ \frac{x_N}{x_1} = \prod_{n=2}^N \frac{x_{n}}{x_{n-1}} = \prod_{n=2}^{N} \left(1 + \frac{1}{4n^2 - 4n}\right)$$ taking logarithm one can obtain $$ \log(x_N) - \log(x_1) = \sum_{n=2}^N \log\left(1 + \frac{1}{4n^2 - 4n}\right) < \frac{1}{4}\sum_{n=2}^N \frac{1}{n^2 - n} = \frac{1}{4} \sum_{n=2}^N \left(\frac{1}{n-1} - \frac{1}{n} \right) $$ hence
$$ \log(x_N) - \log(x_1) < \frac{1}{4} - \frac{1}{4N} \to \frac{1}{4}$$ therefore $$ \lim_{N \to \infty} x_N \leq \frac{1}{4}e^{\frac{1}{4}}$$ Is this correct?
This means that $\frac{1}{4} = x_1 < \lim_{N \to \infty} x_N \leq \frac{1}{4}e^{\frac{1}{4}}$