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?


1 Answer 1


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

$$(-1)^n\frac{\pi^{2n+1}E_{2n}}{(2n)!} = 2^{2n+2}\underbrace{\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{2n+1}}}_{s(2n+1)},$$


$$E_{2n} = (-1)^n 2 \left(\frac{2}{\pi}\right)^{2n+1}s(2n+1)\cdot(2n)!$$

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 $(2)$.

  • $\begingroup$ Awesome. I tried to obtain the asymptotics for $E_{2n}$ using the saddle point method not too long ago and it was pretty messy. This is a very clean approach. $\endgroup$ Nov 25, 2013 at 23:56
  • $\begingroup$ Would you please provide more detail on what you mean by the "partial fraction decomposition" that leads to $(2)$? Normally, partial fraction decomposition is used on the ratio of polynomials. $\endgroup$
    – robjohn
    Dec 20, 2013 at 17:30
  • $\begingroup$ @robjohn Done. In German, we use Partialbruchzerlegung for both (but also Partialbruchentwicklung), apparently the literal translation was not according to common usage. Is "partial fraction development" or "partial fraction expansion" the more common term in English? $\endgroup$ Dec 20, 2013 at 20:17
  • $\begingroup$ @DanielFischer: thanks. This is very reminiscent of the similar identity for $\pi\csc(\pi z)$. (+1) $\endgroup$
    – robjohn
    Dec 20, 2013 at 20:42
  • $\begingroup$ @robjohn It is indeed. One can get each from the other by shifting $z \leadsto z \pm \frac12$, but the fiddling with the correction terms $\frac{1}{a_k}$ resp. $\frac{1}{k}$ one adds for absolute locally uniform convergence gets annoying then (at least, I don't like it), I prefer deriving it from scratch. $\endgroup$ Dec 20, 2013 at 20:46

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