Proof about $z\cot z=1-2\sum_{k\ge1}z^2/(k^2\pi^2-z^2)$ In Concrete Mathematics, it is said that
$$z\cot z=1-2\sum_{k\ge1}\frac{z^2}{k^2\pi^2-z^2}\tag1$$
and proved in EXERCISE 6.73
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\frac{z-k\pi}{2^n}\right)$$
The trigonmetric identity is not hard, but I cannot understand the rest:

It can be shown that term-by-term passage to the limit is justified, hence equation (1) is valid.

How can we conclude that? Thanks for help!

NOTE: This is incomplete. A tighter bound should be produced. Anyone able to do so is free to edit and add it.
For $x$ near the origin, $\cot x \sim \dfrac{1}{x}$. Since $\dfrac{1}{2^n}\to 0 $ we can use this. More precisely,
$$\frac{1}{x}-1<\cot x <\frac{1}{x} $$
Namely, we can dissect
$$z\cot z=\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}+\sum_{k=1}^{2^{n-1}-1}\frac z{2^n}\left(\cot\frac{z+k\pi}{2^n}+\cot\frac{z-k\pi}{2^n}\right)$$
into
$${A_n} = \frac{z}{{{2^n}}}\cot \frac{z}{{{2^n}}} = \cos \frac{z}{{{2^n}}}\frac{{\frac{z}{{{2^n}}}}}{{\sin \frac{z}{{{2^n}}}}} = \cos u\frac{u}{{\sin u}} \to 1$$
$${B_n} = \frac{z}{{{2^n}}}\tan \frac{z}{{{2^n}}} = \frac{z}{{{2^n}}}\sin \frac{z}{{{2^n}}}\frac{1}{{\cos \frac{z}{{{2^n}}}}} = \frac{{u\sin u}}{{\cos u}} \to 0$$
where $u \to 0$.
Then we have
$$\eqalign{
  & \frac{{{2^n}}}{{z - k\pi }} + \frac{{{2^n}}}{{z + k\pi }} - 2 < \left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{{{2^n}}}{{z - k\pi }} + \frac{{{2^n}}}{{z + k\pi }}  \cr 
  & \frac{z}{{z - k\pi }} + \frac{z}{{z + k\pi }} - \frac{z}{{{2^{n - 1}}}} < \frac{z}{{{2^n}}}\left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{z}{{z - k\pi }} + \frac{z}{{z + k\pi }}  \cr 
  & \frac{{2{z^2}}}{{{z^2} - {k^2}{\pi ^2}}} - \frac{z}{{{2^{n - 1}}}} < \frac{z}{{{2^n}}}\left( {\cot \frac{{z + k\pi }}{{{2^n}}} + \cot \frac{{z - k\pi }}{{{2^n}}}} \right) < \frac{{2{z^2}}}{{{z^2} - {k^2}{\pi ^2}}} \cr} $$
