We will use the approximate value of Bernoulli number and Stirling approximation to solve the problem.

• Bernoulli approximate is $\text{B}_{2n}\approx \left( -1\right) ^{n-1}4\sqrt{\pi n} \left( \frac{n}{\pi e} \right) ^{2n}$

• Stirling approximation is $n! \approx \sqrt{2n\pi } \left( \frac{n}{e} \right) ^{n}$

$L=\lim \limits_{n\to \infty }\left[ \frac{\left( 2n\right) ! }{\left| \text{B}_{2n}\right| } \right] ^{\frac{1}{2n} }=\lim \limits_{n\to \infty }\left[ \frac{2\sqrt{n\pi } \left( \frac{2n}{e} \right) ^{2n}}{4\sqrt{n\pi } \left( \frac{n}{\pi e} \right) ^{2n}} \right] ^{\frac{1}{2n} } \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ = \lim \limits_{n\to \infty }\left[ \frac{\left( 2\pi \right) ^{2n}}{2} \right] ^{\frac{1}{2n} }=2\pi. \lim \limits_{n\to \infty }\left[ \frac{1}{2} \right] ^{\frac{1}{2n} }=\boxed {2\pi \approx 6.283185}$

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Edit

Another solution using the relation between zeta function and Bernoulli numbers $\text{B}_{2n}=\frac{2\left( -1\right) ^{n-1}\left( 2n\right) ! }{\left( 2\pi \right) ^{2n }} \zeta \left( 2n\right)$ So the limit will be L=\underbrace{\lim \limits_{_{n\to +\infty }}\left[ \frac{\left( 2\pi \right) ^{2n}}{2} \right] } \limits_{L_{1}}\times \underbrace{\lim \limits_{n\to +\infty }\left[ \zeta \left( 2n\right) \right] ^{\frac{1}{2n} }} \limits_{L_{2}} From the first part we see that $L_{1}=2\pi$ . Now we will prove that $L_{2}=1$

The proof

$\left[ \left( 1\right) ^{\frac{1}{2n} }\leq \left( \zeta \left( 2n\right) \right) ^{\frac{1}{2n} }\leq \left[ 1+\frac{1}{2n-1} \right] ^{\frac{1}{2n} }\right] _{n\to \infty }=1$ $\text{Q.E.D}$