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!

This identity is also proven in this answer, but the limit of the trigonometric identity is a cute trick, too.
Concrete Mathematics claim:
For the limit claimed in Concrete Mathematics, we need a few things.
First, by inspecting the graph of $\frac{1-x\cot(x)}{x^2}$ for $-\frac{3\pi}{4}\le x\le\frac{3\pi}{4}$, we have
$$
\left|\frac1x-\cot(x)\right|\le|x|\tag{1}
$$
Next, the Mean Value Theorem says
$$
\begin{align}
|\cot(\delta+x)+\cot(\delta-x)|
&=|\cot(x+\delta)-\cot(x-\delta)|\\
&\le2\delta\sup_{[x-\delta,x+\delta]}\csc^2(\xi)\\
&\le\color{#C00000}{8\delta\,\csc^2(x)}\\
&\le\color{#C00000}{2\pi^2\delta/x^2}\tag{2}
\end{align}
$$
if $\color{#C00000}{2\delta\le|x|\le\frac{\pi}{2}}$.
Finally, note that since $0\le k< 2^{n-1}$, $0\le\frac{k\pi}{2^n}<\frac{\pi}{2}$
Using $(1)$, we get
$$
\begin{align}
&\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)-\left(\frac{z}{z+k\pi}+\frac{z}{z-k\pi}\right)\right|\\
&\le2\left|\frac{z}{2^n}\right|\frac{|z|+k\pi}{2^n}\tag{3}
\end{align}
$$
Using $(2)$, we get, for $2z\le k\pi$,
$$
\begin{align}
\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)\right|
&\le2\pi^2\left|\frac{z^2}{2^{2n}}\right|\left(\frac{2^n}{k\pi}\right)^2\\
&\le2\pi^2\left(\frac{z}{k\pi}\right)^2\tag{4}
\end{align}
$$
Estimate $(3)$ is used to control the difference between the series for small $k$, and $(4)$ to control the remainder in the sum of the cotangents for large $k$.
Pick an $\epsilon>0$, and find $m$ large enough so that $2z\le m\pi$ and
$$
\sum_{k=m}^\infty\frac{1}{k^2}\le\epsilon\tag{5}
$$
Then we have the following estimate for the tail of the sum
$$
\sum_{k=m}^\infty\frac{z^2}{k^2\pi^2-z^2}\le\frac43z^2\epsilon\tag{6}
$$
Combining $(4)$ and $(5)$ yields
$$
\sum_{k=m}^{2^{n-1}-1}\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)\right|\le2z^2\epsilon\tag{7}
$$
Summing $(3)$ gives
$$
\begin{align}
&\sum_{k=1}^{m-1}\left|\frac{z}{2^n}\left(\cot\left(\frac{z+k\pi}{2^n}\right)+\cot\left(\frac{z-k\pi}{2^n}\right)\right)-\left(\frac{z}{z+k\pi}+\frac{z}{z-k\pi}\right)\right|\\
&\le2\left|\frac{z}{2^n}\right|\frac{m|z|+m^2\pi/2}{2^n}\tag{8}
\end{align}
$$
Just choose $n$ big enough so that $(8)$ and $\displaystyle\left|\frac z{2^n}\cot\frac z{2^n}-\frac z{2^n}\tan\frac z{2^n}-1\right|$ are each less than $\epsilon$ and we get that the term-by-term absolute difference is less than
$$
\left(\frac{10}{3}z^2+2\right)\epsilon\tag{9}
$$
