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Recall that the Laurent series of $\displaystyle\frac{1}{e^z-1}$ near $z=0$ is given as

$$\frac{1}{e^z-1}=\frac1z-\frac12+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{(2k)!}B_k z^{2k-1},$$

where the $B_k$ are the Bernoulli numbers. (This definition of the Bernoulli numbers is slightly different with that from the Wikipedia, but this definition is just the nonzero terms with all positive sign.)

I want to prove that $\displaystyle\lim_{k \to \infty} B_k = \infty$, but I have no idea. How do I have to prove these kinds of statements?

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  • $\begingroup$ One way to prove it is to find a different formula for the coefficients via a partial fraction decomposition. The partial fraction decomposition of $\pi \cot (\pi z)$ is pretty famous and closely related. $\endgroup$ – Daniel Fischer Nov 26 '19 at 14:04
  • $\begingroup$ Check out a classical result of Euler relating the zeta values at $2n$ with Bernoulli numbers $B_{2n}$, which up to sign is your "$B_n$". $\endgroup$ – Pythagoras Nov 27 '19 at 20:53
  • $\begingroup$ @DanielFischer I derived that $\pi \cot \pi z= 1/z - \sum _{k=1} ^\infty B_k (2 \pi )^{2k} z^{2k-1} / (2k)!$ but I can't see nothing.. $\endgroup$ – user302934 Nov 27 '19 at 21:26
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    $\begingroup$ The partial fraction decomposition of $\pi \cot (\pi z)$ is $$\frac{1}{z} + \sum_{n \in \mathbb{Z}\setminus \{0\}} \biggl(\frac{1}{z-n} + \frac{1}{n}\biggr) = \frac{1}{z} + 2z \sum_{n = 1}^{\infty} \frac{1}{z^2 - n^2}\,.$$ In the last sum, for $\lvert z\rvert < 1$, expand $\frac{1}{z^2-n^2}$ into a geometric series, then change the order of summation. You get a form of the Laurent series in which the coefficients are given in terms of $\zeta(2k)$ and not in terms of the $B_k$, and thus a relation between $B_k$ and $\zeta(2k)$. (That's the classical result of Euler that Pythagoras refers to.) $\endgroup$ – Daniel Fischer Nov 27 '19 at 21:36
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    $\begingroup$ And that you already found. Rewrite that relation so that $B_n$ stands alone on one side. The other side can easily be shown to tend to $\infty$ (if you know Stirling, you can use that, but much weaker estimates suffice). $\endgroup$ – Daniel Fischer Nov 27 '19 at 21:39
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To avoid confusion, it might be a better idea to keep a more conventional way to define the Bernoulli numbers, namely by the generating function $$\frac z{e^z-1}=\sum_{n=0}^\infty B_n\frac {z^n}{n!}.$$ The function $\frac z{e^z-1}-1+\frac 12z$ being even, one has $$\frac z{e^z-1}=1-\frac 12z+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}z^{2k}.$$ Then your question boils down to asserting that $$\lim_{k\rightarrow\infty}(-1)^{k+1}B_{2k}=\infty.$$ This is just a classical result of Euler relating the zeta values to Bernoulli numbers:

Theorem $B_{2n}=\frac {(-1)^{n+1}2(2n)!}{(2\pi)^{2n}}\zeta(2n)$

For the formula, you may look up proof or use the other suggested comments. Granting this, note that $\zeta(2n)>1$, so one gets that $$|B_{2n}|>\frac {2(2n)!}{(2\pi)^{2n}},$$ which goes to $\infty$ by trivial comparison test.

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Idea, too long for a commentary: $$f(z) = \frac{1}{e^z - 1} - \frac1z + \frac12$$ has a removable singularity at $z = 0$. Defining $f(0) = 0$, you can apply the Cauchy integral formulas (Cauchy's differentiation formula), writing $f^{(n)}(0)$ as a line integral along a small circle $|z| = \epsilon$ and estimating a lower bound of the absolute value of the integral.

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