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

Question

Recently I encountered this problem

Show that $$\sum_{n = 1}^{\infty}\frac{1}{x_{n}^{2}} = \frac{1}{10}$$ where $x_{n}$ is the $n^{\text{th}}$ positive root of $\tan x = x$.

I found many threads on MSE regarding solutions to $\tan x = x$ like this and this which suggest that the series $\sum 1/x_{n}^{2}$ is convergent if we compare it with $\sum 1/n^{2}$. But at the same time I have no idea how we can sum this up. Please help with suggestions or an answer.

Update: I further tried to relate it with the sum $\sum 1/n^{2} = \pi^{2}/6$ and we can rewrite it as $\sum 1/(n^{2}\pi^{2}) = 1/6$ so that if $y_{n}$ is $n^{\text{th}}$ positive root of $\sin x = 0$ then $\sum 1/y_{n}^{2} = 1/6$. I believe this has got to do with the product $$\frac{\sin x}{x} = \prod_{n = 1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}\pi^{2}}\right)$$ where we can see the factors corresponding to actual roots. We can now compare the coefficients of $x^{2}$ on both sides to get $\sum 1/y_{n}^{2} = 1/6$. I don't see any product related to equation $\tan x = x$.

Answer 1

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 $$ f(z) = \left(1 - \frac{z^2}{3!} + \frac{z^4}{5!}\right) - \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!}\right) + O(z^6) = \frac{z^2}{3} - \frac{z^4}{30} + O(z^6)\tag{*1}$$ 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.

Answer 2

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.

Answer 3

This is not an exact answer (you got some which are very good); so, forgive me if it does not fit your needs.

In a previous post there exist some real $a >0$ such that $\tan{a} = a$ , I showed that the $k^{th}$ solution of $\tan(x)=x$ can be approximated by $$x_k\simeq(2k+1) \frac {\pi}{2}-\frac{2}{\pi (2 k+1)}$$ If the second term is ignored $x_k$ is just the value of $\tan(x)$ is undefined and then $$S_1=\sum_{n = 1}^{\infty}\frac{1}{x_{n}^{2}} \simeq \frac{\pi ^2-8}{2 \pi ^2} \simeq 0.0947153$$ Using both terms of the approximation lead to something more complex and $$S_2=\sum_{n = 1}^{\infty}\frac{1}{x_{n}^{2}} \simeq \frac{1}{4} \left(-\frac{16 \pi ^2}{\left(\pi ^2-4\right)^2}+\tan (1)+\sec ^2(1)\right) \simeq 0.0998439$$

If, instead, we use $$x_k \simeq \frac{1}{4}\Big((2k+1)\pi+\sqrt {(2k+1)^2 \pi^2-16}\Big)$$ from which were derived the previous approximations for large values of $k$, the summation cannot be any more computed analytically ans its numerical value is $0.100087$.