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Question is in the title.

We have:

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

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    $\begingroup$ $1<\zeta(2n)=(-1)^{n+1}B_{2n}(2\pi)^{2n}/(2\times(2n)!)$ (+ Stirling for $(2n)!$) $\endgroup$ – user8268 May 6 '14 at 12:21
  • $\begingroup$ Given a generating function for a sequence, there is a method that relates the asymptotics of the sequence to properties of the function on the boundary of the disk of convergence. $\endgroup$ – GEdgar May 6 '14 at 12:23
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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.

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  • $\begingroup$ Thanks. This is way over my head! I just found out that it can be shown using the Riemann Zeta function: en.wikipedia.org/wiki/Bernoulli_number#Asymptotic_approximation Can this be made rigorous? (only taking the first term from the Zeta series). $\endgroup$ – Superbus May 7 '14 at 3:34
  • $\begingroup$ This is essentially what I have above, consult this MSE link for more details. Some may prefer not to have the dependence on the zeta function values. You may however take more terms from Stirling's approximation as well as contributions from additional poles. (Signing off for the day.) $\endgroup$ – Marko Riedel May 7 '14 at 3:44

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