TITLE: Sum of the squares of the reciprocals of the fixed points of the tangent function
QUESTION [27 upvotes]: The sum of the squares of the reciprocals of the positive fixed points of the tangent function is $1/10$.
I've seen this proved by means of residues, but I don't remember the details.
I've also heard it asserted that it that it can be done by means of Green's functions.
What proofs of this fact are published or otherwise known?
PS: Maybe it is of interest to note that the positive fixed points of the tangent function are also the abscissas of the extreme values of the function $x\mapsto\dfrac{\sin x} x.$

REPLY [3 votes]: I answered this similar question a while ago, so I thought I would answer this question in the same manner, with more detail.

Residue at $\boldsymbol{z=0}$
Compute the series near $x=0$ for $\tan(x)$ as the ratio of the series for $\sin(x)$ and $\cos(x)$:
$$
\begin{align}
\tan(x)
&=\frac{x-\frac16x^3+\frac1{120}x^5+O\!\left(x^7\right)}{1-\frac12x^2+\frac1{24}x^4+O\!\left(x^6\right)}\\
&=\left(x-\frac16x^3+\frac1{120}x^5+O\!\left(x^7\right)\right)\left(1+\frac12x^2+\frac5{24}x^4+O\!\left(x^6\right)\right)\\
&=x+\frac13x^3+\frac2{15}x^5+O\!\left(x^7\right)\tag1
\end{align}
$$
compute the series near $x=0$ for $\frac1{\tan(x)-x}$ using $(1)$:
$$
\begin{align}
\frac1{\tan(x)-x}
&=\frac1{\frac13x^3+\frac2{15}x^5+O\!\left(x^7\right)}\\
&=\frac3{x^3}\left(1-\frac25x^2+O\!\left(x^4\right)\right)\\
&=\frac3{x^3}-\frac6{5x}+O(x)\tag2
\end{align}
$$
Therefore,
$$\newcommand{\Res}{\operatorname*{Res}}
\Res_{z=0}\left(\frac1{\tan(z)-z}\right)=-\frac65\tag3
$$

Residue at Other Singularities
At any point $z_k\ne0$ where $\tan(z_k)=z_k$,
$$
\begin{align}
\Res_{z=z_k}\frac1{\tan(z)-z}
&=\lim_{z\to z_k}\frac{z-z_k}{\tan(z)-z}\\
&=\lim_{z\to z_k}\frac1{\sec^2(z)-1}\\
&=\frac1{\tan^2(z_k)}\\
&=\frac1{z_k^2}\tag4
\end{align}
$$
Enumerate the positive roots of $\tan(z)=z$ as $\{z_k\}_{k=1}^\infty$. Note that $\{-z_k\}_{k=1}^\infty$ are the negative roots of $\tan(z)=z$.

Behavior Near $\boldsymbol{\infty}$
Since $\tan(n\pi+iy)=i\tanh(y)$ for $n\in\mathbb{Z}$, we get that
$$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}}
\begin{align}
|\tan(n\pi+iy)|
&=|\tanh(y)|\\
\le1\tag5
\end{align}
$$
and 
$$
\begin{align}
|\tan(x+iy)|
&=\left|\frac{e^{ix-y}-e^{-ix+y}}{e^{ix-y}+e^{-ix+y}}\right|\\
&\le\left|\frac{\cosh(y)}{\sinh(y)}\right|\\
&=|\coth(y)|\tag6
\end{align}
$$
Thus, on the path $\max(|\Re(z)|,|\Im(z)|)=n\pi$, $n\in\mathbb{Z}$, we have
$$
|\tan(z)|\le\coth(\pi)\tag7
$$
Thus, on that path
$$
\begin{align}
\frac1{\tan(z)-z}
&=\frac1{O(1)-z}\\
&=-\frac1z+O\!\left(\frac1{z^2}\right)\tag8
\end{align}
$$
Thus, the integral counter-clockwise around that path is $-2\pi i$.

The Sum
Putting $(3)$, $(4)$, and $(8)$ together, we get
$$
-2\pi i=2\pi i\left(-\frac65+2\sum_{k=1}^\infty\frac1{z_k^2}\right)\tag9
$$
which gives
$$
\sum_{k=1}^\infty\frac1{z_k^2}=\frac1{10}\tag{10}
$$