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} $