Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial? Stirling's approximation of the factorial for even numbers is given by
$$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$
Further, the Euler numbers grow quite rapidly for large indices as they have the following lower bound
$$    |E_{2 n}| > 8 \sqrt { \frac{n}{\pi} } \left(\frac{4 n}{ \pi e}\right)^{2 n}=
2\left(\frac{2}{ \pi }\right)^{2 n+1} \sqrt { 4n\pi} \left(\frac{2 n}{  e}\right)^{2 n}. \tag{2}$$
Why do these formulas look so similar? Or is $(2)$ just a way to say that Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?
 A: There's more to it, we have
$$\frac{1}{\cos z} = \sum_{n=0}^\infty (-1)^n \frac{E_{2n}}{(2n)!}z^{2n}.\tag{1}$$
Since $\dfrac{1}{\cos z}$ has poles in $(k+\frac12)\pi$ for $k\in\mathbb{Z}$ and is holomorphic everywhere else, the series $(1)$ has a radius of convergence of $\dfrac{\pi}{2}$. The Cauchy-Hadamard formula now says
$$\limsup_{n\to\infty} \sqrt[n]{\frac{\lvert E_{2n}\rvert}{(2n)!}} = \frac{2}{\pi}.$$
More precisely, from the partial fraction development of $\dfrac{\pi}{\cos \pi z}$ (see below) we obtain
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.\tag{2}$$
Replacing $z$ with $\pi z$ in $(1)$ and multiplying with $\pi$ yields
$$\frac{\pi}{\cos \pi z} = \sum_{n=0}^\infty (-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!}z^{2n},\tag{3}$$
and comparing coefficients with $(2)$ yields
$$(-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!} = 2^{2n+2}\underbrace{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}}_{s(2n+1)},$$
or
$$E_{2n} = (-1)^n 2 \left(\frac{2}{\pi}\right)^{2n+1}s(2n+1)\cdot(2n)!$$
Since $s(2n+1) \to 1$ for $n\to \infty$, we have
$$\lvert E_{2n}\rvert \sim 2\left(\frac{2}{\pi}\right)^{2n+1}\cdot(2n)!$$

$\cos \pi z$ has zeros in $a_k = k + \frac12,\: k \in \mathbb{Z}$ and nowhere else. All zeros are simple. The residue of
$$f(z) := \frac{\pi}{\cos \pi z}$$
in $a_k$ is therefore
$$\operatorname{Res}\left(f;a_k\right) = \frac{\pi}{\frac{d}{dz}(\cos \pi z)\bigl\lvert_{a_k}} = \frac{1}{-\sin \pi a_k} = (-1)^{k+1}.$$
The series
$$g(z) = \sum_{k\in \mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right)$$
is absolutely and locally uniformly convergent, and therefore $g$ is an entire meromorphic function with the same principal parts as $f$, thus $f-g$ is an entire holomorphic function.
$$g'(z) = \sum_{k\in\mathbb{Z}} \frac{(-1)^k}{(z-a_k)^2}$$
is easily seen to be an entire meromorphic function with period $2$ and the property that $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$. The same holds for
$$f'(z) = \frac{\pi^2\sin \pi z}{\cos^2\pi z},$$
and furthermore $f'$ and $g'$ have the same poles and principal parts. Hence $f' - g'$ is an entire holomorphic function with period $2$ and $\lim\limits_{\lvert \operatorname{Im} z\rvert\to \infty} \lvert f'(z) - g'(z)\rvert = 0$ uniformly in $\{ z : \lvert \operatorname{Re} z\rvert \leqslant 1\}$, thus bounded, ergo constant, and the vanishing of $f'-g'$ for growing imaginary part determines that constant as $0$.
Therefore $f-g$ is constant, and since $g(0) = 0$, we have
$$\frac{\pi}{\cos \pi z} = \pi + \sum_{k\in\mathbb{Z}} (-1)^{k+1}\left(\frac{1}{z-a_k} + \frac{1}{a_k}\right).$$
Grouping, for $k \geqslant 0$, the terms for the two indices $k$ and $-(k+1)$, noting $a_{-(k+1)} = -a_k$, we obtain for $\lvert z\rvert < \frac12$
$$\begin{align}
\frac{\pi}{\cos\pi z} &= \pi + \sum_{k=0}^\infty (-1)^k \left(\frac{1}{a_k-z} + \frac{1}{a_k+z} - \frac{2}{a_k}\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\left(\frac{a_k^2}{a_k^2-z^2}-1\right)\\
&= \pi + \sum_{k=0}^\infty \frac{(-1)^k2}{a_k}\sum_{n=1}^\infty \left(\frac{z}{a_k}\right)^{2n}\\
&= \pi + \sum_{n=1}^\infty \sum_{k=0}^\infty \frac{(-1)^k2}{a_k^{2n+1}}z^{2n}\\
&= \pi + \sum_{n=1}^\infty \left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}\right)2^{2n+2}z^{2n}.
\end{align}$$
Writing $\pi$ as the Leibniz series  $\frac{\pi}{4} = 1 - \frac{1}{3} +\frac{1}{5} - \dotsb$ then yields $(2)$.
