Why do the even Bernoulli numbers grow so fast?

Question

Question is in the title.

We have:

$$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$

Answer 1

Recall the generating function $$f(z) = \sum_{q\ge 0} B_q \frac{z^q}{q!} = \frac{z}{e^z-1}.$$

We can treat it as if it were rational as the poles are countable with no accumulation point, so we may use the techniques of extracting coefficients from rational generating functions, with the asymptotics being given by inverse powers of the dominant poles.

The pole at $z=0$ is cancelled by the factor $z$ in the numerator. The nearest two poles are at $\pm 2\pi i,$ both at the same distance from zero ($2\pi$).

Computing the residues we obtain $$\mathrm{Res}(f(z); z=\pm 2\pi i) = \pm 2\pi i$$ and from this we get the singular decomposition $$\frac{2\pi i}{z-2\pi i} - \frac{2\pi i}{z+2\pi i}.$$ Turn this into geometric series to get $$\frac{1}{z/(2\pi i)-1} - \frac{1}{z/(2\pi i)+1} = - \frac{1}{1-z/(2\pi i)} - \frac{1}{1+z/(2\pi i)} \\= -\sum_{q\ge 0} \frac{z^q}{(2\pi i)^q} -\sum_{q\ge 0} (-1)^q \frac{z^q}{(2\pi i)^q} = -2 \sum_{q\ge 0} \frac{z^{2q}}{(2\pi i)^{2q}} \\ = -2 \sum_{q\ge 0} (-1)^q \frac{z^{2q}}{(2\pi)^{2q}} = 2 \sum_{q\ge 0} (-1)^{q+1} \frac{z^{2q}}{(2\pi)^{2q}}.$$

Extracting coeffcients and since $B_{2n} = (2n)! [z^{2n}] f(z)$ we find that $$B_{2n} \sim (2n)! \times 2 \times \frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$ Using Stirling's formula this becomes $$\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} \times 2 \times \frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$ This is $$4 \times (-1)^{n+1} \times \sqrt{\pi n} \left(\frac{n}{\pi e}\right)^{2n}.$$

Addendum. To justify rigorously the above asymptotics we need to verify that there isn't an additional entire component not accounted for in the complete singular decomposition.

To do this introduce the sum $$g(w) = \sum_{q\ge 1}\left(\frac{1}{w/q-1}-\frac{1}{w/q+1}\right) = 2 \sum_{q\ge 1} \frac{1}{w^2/q^2-1}.$$

This can be evaluated using a standard technique from complex variables and setting $$G(z) = 1/(w^2/z^2-1)\times\pi\cot(\pi z)$$ we have $$g(w) = - \left( \mathrm{Res}(G(z); z=0) +\mathrm{Res}(G(z); z=w) +\mathrm{Res}(G(z); z=-w) \right).$$ Computing the residues we get $$g(w) = \pi w \cot(\pi w).$$ Now we claim that $$f(z) = -\frac{1}{2}z + \sum_{q\ge 1} \left(\frac{2\pi i q}{z-2\pi i q}-\frac{2\pi i q}{z+2\pi i q}\right).$$ The sum is $g(z/(2\pi i))$ so we get $$-\frac{1}{2}z + \frac{z}{2i} \cot\left(\frac{z}{2i}\right) = -\frac{1}{2}z + \frac{z}{2i} \times i \times \frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}} \\= \frac{1}{2} z \left(-1 + \frac{e^z+1}{e^z-1}\right) = \frac{1}{2}z \frac{2}{e^z-1} = \frac{z}{e^z-1}.$$

This shows rigorously that the entire part is $$-\frac{1}{2}z$$ and hence makes no contribution to $B_{2n}$ where $n\ge 1$ and the above singular decomposition is justified.