Bernoulli numbers tend to infinity

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?

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.