Using the result $\alpha_{2n+1} = \frac{2(2^{2n+2}-1)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2)$ and the fact that $\lim_{s \to +\infty} \zeta(s) = 1$, deduce an equivalent of $\alpha_{2n+1}$ as $n$ tends to infinity.
Using the result $\alpha_{2n+1} = \frac{2(2^{2n+2}-1)(2n+1)!}{\pi^{2n+2}} \zeta(2n+2)$ and the fact that $\lim_{s \to +\infty} \zeta(s) = 1$, deduce an equivalent of $\alpha_{2n+1}$ as $n$ tends to infinity.