Show that $\sum\limits_n1/x_{n}^{2} = 1/10$ where $x_{n}$ is the $n^{\text{th}}$ positive root of $\tan x = x$ [duplicate]

Notice for non-zero $z$, we have

$$\tan z = z \quad\iff\quad f(z) = \frac{\sin z}{z} - \cos z = 0$$

Since $\sin z$ and $\cos z$ are entire functions of order $1$ at infinity, so does $f(z)$.

For small $z$, we have $$ f(z) = \left(1 - \frac{z^2}{3!} + \frac{z^4}{5!}\right) - \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!}\right) + O(z^6) = \frac{z^2}{3} - \frac{z^2}{30} + O(z^4)\tag{*1}$$ This means $z = 0$ is a double root for $f(z)$.

$f(z)$ is clearly an even function. Its non-zero roots will come in pairs. Let $\pm \alpha_n, n = 1, 2,\ldots$ be the non-zero roots ordered by $0 < |\alpha_1| \le |\alpha_2| \le \ldots$. By Hadamard factorization theorem, $f(z)$ have following factorization:

$$f(z) = z^2 e^{g(z)} \prod_{n=1}^\infty \left(\left(1-\frac{z}{\alpha_n}\right)e^{\frac{z}{\alpha_n}}\right) \left(\left(1+\frac{z}{\alpha_n}\right)e^{-\frac{z}{\alpha_n}}\right) = z^2 e^{g(z)} \prod_{n=1}^\infty \left(1 - \frac{z^2}{\alpha_n^2}\right) $$ where $g(z)$ is a polynomial of degree at most $1$.

Once again, since $f(z)$ is an even function, $\deg g(z) \le 1 \implies g(z)$ is a constant. Compare this with $(*1)$, we get

$$f(z) = \frac{z^2}{3} \prod_{n=1}^\infty \left(1-\frac{z^2}{\alpha_n^2}\right)$$

Assume one can show that all the $\alpha_n$ lies on real axis and $|\alpha_n|$ increases so quickly such that $\displaystyle\;\sum_{n=1}^\infty \frac{1}{a_n^2}\;$ converges. It is then legal to expand the infinite product and obtain $$f(z) = \frac{z^2}{3} \left( 1 - \left(\sum_{n=1}^\infty \frac{1}{\alpha_n^2}\right) z^2 + O(z^4) \right)$$ Compare this with $(*1)$ again, we get $\displaystyle\;\sum_{n=1}^\infty \frac{1}{\alpha_n^2} = \frac{1}{10}\;$.

Update

To see why all $\alpha_n$ lies on the real axis, let $z = x + iy$ be any non-real root of $\tan z = z$, we have

$$\frac{\sin z}{z} = \cos z \iff \frac{e^{iz} - e^{-iz}}{iz} = e^{iz} + e^{-iz} \iff e^{iz}(1-iz) = e^{-iz}(1+iz)$$ Taking absolute value and square on both sides of last expression, it is equivalent to $$ e^{-2y}((1+y)^2 + x^2) = e^{2y}((1-y)^2 + x^2) \iff \sinh(2y)(x^2 + y^2 + 1) - 2y\cosh(2y) = 0 $$ This leads to $$x^2 = \frac{1}{\sinh^2(2y)} - (y-\coth(2y))^2 = - (y - \tanh(y))(y-\coth(y)) \tag{*2}$$ The equation $y = \coth(y)$ has a unique root $\mu \approx 1.199678640257734$ on postive real axis. When $|y| > \mu$, RHS of $(*2)$ is negative and there is no real solution for $x$.

A plot of RHS of $(*2)$ over the interval $[-\mu,\mu]$ show that it is bounded above by $0.1$. If $\tan z = z$ has any non-real root, that root need to fall within a small rectangle $$\big\{\; x + iy : |x| \le \sqrt{0.1}, |y| \le \mu\;\big\}$$ A plot of $f(z)$ over this small rectangle indicate there isn't any non-real root. From this, we can conclude all $\alpha_n$ lies on the real axis.

Finally, about the issue whether $\displaystyle\;\sum_{n=1}^\infty \frac{1}{\alpha_n^2}$ converges or not. One can superpose the plot of $z$ vs $z$ over the plot of $\tan z$ vs. $z$. One will observe $$n \pi < \alpha_n < (n + \frac12)\pi,\quad \forall N$$ This immediately leads to $$\sum_{n=1}^\infty \frac{1}{\alpha_n^2} < \frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2} = \frac16 < \infty$$ and the sum over $\displaystyle\;\frac{1}{\alpha_n^2}\;$ does converge.

Edit: Something like this is already mentioned in the comments.

Not a full answer but a nice observation. Let $$f(z) = \prod_{k=1}^{\infty}\left(1-\frac{z^2}{x_n^2}\right).$$ Then $f$ is entire, $f(0)=1$ and the (simple) roots of $f$ are exactly $\pm x_k$ for $k\geq 1$. So if you can show that $$f(z)=\frac{3(\sin(z)-z \cos(z))}{z^3}=1-\frac{1}{10}z^2+O(z^4)$$ then you're done.