If you are looking for a trucated series, you could start from $$\tan(z)=z+\frac{z^3}{3}+\frac{2 z^5}{15}+\frac{17 z^7}{315}+\frac{62 z^9}{2835}+O\left(z^{11}\right)$$ which makes $$\cot(z)=\frac 1{z+\frac{z^3}{3}+\frac{2 z^5}{15}+\frac{17 z^7}{315}+\frac{62 z^9}{2835}+O\left(z^{11}\right)}=\frac 1z \frac 1{1+\frac{z^2}{3}+\frac{2 z^4}{15}+\frac{17 z^6}{315}+\frac{62 z^8}{2835}+O\left(z^{10}\right)}$$ and perform long division to get $$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}-\frac{2 z^5}{945}-\frac{z^7}{4725}+O\left(z^9\right)$$ If you want the infinite series consider that $$\cot(z)=\frac 1 {\tan(z)}=f(z)=\sum_{i=0}^\infty a_iz^{i-1}$$ what you can rewrite as $$1=\tan(z)\sum_{i=0}^\infty a_iz^{i-1}$$ that is to say $$1=\left(\sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} z^{2n-1}\right)\times \sum_{i=0}^\infty a_iz^{i-1}$$ For simplicity, let us define $$b_n=\frac{B_{2n} (-4)^n (1-4^n)}{(2n)!}$$ in order to solve $$1=\sum^{\infty}_{n=1}b_nz^{2n-1} \times \sum_{i=0}^\infty a_iz^{i-1}$$ Developing, we get $$1=a_0 b_1+a_1 b_1 z+ (a_2 b_1+a_0 b_2)z^2+ (a_3 b_1+a_1 b_2)z^3+ (a_4 b_1+a_2 b_2+a_0 b_3)z^4+ (a_5 b_1+a_3 b_2+a_1 b_3)z^5+ (a_6 b_1+a_4 b_2+a_2 b_3+a_0 b_4)z^6+(a_7 b_1+a_5 b_2+a_3 b_3+a_1 b_4)z^7 +\cdots$$ Now,we need to solve, for the $a_i$'s the equations $$a_0 b_1=1$$ $$a_1 b_1=0$$ $$a_2 b_1+a_0 b_2=0$$ $$a_3 b_1+a_1 b_2=0$$ $$a_4 b_1+a_2 b_2+a_0 b_3=0$$ $$a_5 b_1+a_3 b_2+a_1 b_3=0$$ $$a_6 b_1+a_4 b_2+a_2 b_3+a_0 b_4=0$$ $$a_7 b_1+a_5 b_2+a_3 b_3+a_1 b_4=0$$ This does not make much problem (using successive eliminations for example).

This leads to the infinite series $$\cot(z)=\sum_{n=0}^\infty (-1)^n\frac{ 2^{2 n}\, B_{2 n} }{(2 n)!}z^{2 n-1}$$