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!

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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 $$\left(\frac{10}{3}z^2+2\right)\epsilon\tag{9}$$

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Well, but I wonder why term-by-term passage is justified. Peter's proof is just using squeeze theorem, but I don't know whether there's a systematical theory to explain it, maybe uniform convergence? – Frank Science Jun 27 '12 at 13:58
@FrankScience: sorry for taking so long, but I hope this answers your question. – robjohn Jun 28 '12 at 15:32

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}

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Summing up equation (5) and use squeeze theorem. Thanks. Incidentally, your $\LaTeX$ with nested $\lbrace\cdots\rbrace$ is awful, so I've edited your post. – Frank Science Jun 18 '12 at 4:45
@FrankScience OK, but don't crop my answer. Thanks. – Pedro Tamaroff Jun 18 '12 at 9:15
@FrankScience I don't nest the parenthesis, MathType does. It'd have been more prudent to ask, you know. – Pedro Tamaroff Jun 18 '12 at 16:51
I suggest that you be famailiar with $\LaTeX$. I find it more convenient to use $\LaTeX$ directly if you can spend half an hour learning it. Some tutorials are useful, say, lshort. – Frank Science Jun 19 '12 at 2:01
@FrankScience I know some $\LaTeX$ but I can't waste $10$ mins coding it, thanks anyways. – Pedro Tamaroff Jun 19 '12 at 2:10