Consider the sum $S(n)=\sum\dfrac{1}{x^2}$ where summation is performed over all the positive real solutions of the equation $\dfrac{\tan{x}}{x}=n$.
If it's given that $S(n)=1$, $n\in\mathbb{Q}$, find $n$.
My attempt:
We're interested in the roots of
$$\frac{\tan x}{x}=k$$
Then, performing Taylor expansion of
$$\sin x = kx\cos x$$
$$x- \frac{(x^6)}{6}+ ...... = kx(1- \frac{x^2}{2}+........$$
I am stuck here. Any hints will be appreciated.
Thanks
Consider the Taylor series and Hadamard factorization of the function $$ f_n(z)=\frac{\sin z}z-n\cos z,\tag1 $$ which are: $$ \begin{align} f_n(z)&=1-\frac16z^2+\dots-n\left(1-\frac12z^2+\dots\right)\\ &=(1-n)-\left(\frac16-\frac n2\right)z^2+\dots\tag2 \end{align} $$ and $$ \begin{align} f_n(z)&=f_n(0)\prod_{\rho}^{f_n(\rho)=0}\left(1-\frac z{\rho}\right)e^{z/\rho}\\ &=f_n(0)\prod_{\rho_*}^{f_n(\rho_*)=0}\left(1-\frac {z^2}{\rho_*^2}\right)\\ &=(1-n)\left(1-z^2\sum_{\rho_*}^{f_n(\rho_*)=0}\frac1{\rho_*^2}+\dots\right),\tag3 \end{align} $$ respectively, where $n\ne1$ is assumed and we used the fact that all roots $\rho$ are paired $f_n(\pm\rho)=0$. Due to this fact we can effectively use one root $\rho_*$ from each pair. For convenience we may assume $0\le\operatorname{Arg}(\rho_*)<\pi$.
Comparing the coefficients at $z^2$ in $(2)$ and $(3)$ one obtains the equality: $$%\sideset{}' (1-n)\sum_{\rho_*}^{f_n(\rho_*)=0}\frac1{\rho_*^2}=\frac16-\frac n2.\tag4 $$
The case $n=1$ can be treated similarly by considering the function $F(z)=\frac1{z^2}f_1(z)$ which leads to the value $S(1)=\frac1{10}$.
A fine detail here is the domain of summation. Whereas the sum in question runs over the real roots, the sum in $(4)$ runs over all complex roots of the function $f_n(z)$. The two sets are not necessarily the same, as we will immediately see. Indeed: $$\begin{align} &f_n(z)=0\\ &\implies |\sin z-nz\cos z|^2=0\\ &\stackrel{z=x+iy}\implies (\sin x-nx\cos x)^2+(\sinh y-ny\cosh y)^2 +n^2x^2y^2\left(\frac{\sinh^2y}{y^2}-\frac{\sin^2x}{x^2}\right)=0. \end{align} $$ Obviously the last equality can hold only if $x=0$ or $y=0$: $$\begin{align} x=0:&\quad \tanh y=n y;\tag{5}\\ y=0:&\quad \tan x=n x.\tag{6}\\ \end{align} $$ Whereas the equation $(6)$ delivers all non-zero real roots of the function $f_n(z)$ the equation $(5)$ results in an additional pair of imaginary roots provided that $0<n<1$ (otherwise $(5)$ has no non-zero real solutions).
Let the imaginary root of $f_n(z)$ be (if it exists) $iy_n$, where $y_n$ is the non-zero real solution of $(5)$. Then $(4)$ will read: $$ -\frac{\mathbb1_{0<n<1}}{y_n^2}+S(n)=\frac{\frac16-\frac n2}{1-n}.\tag7 $$
Coming back to the original problem this means that the assumption $n\not\in(0,1)$ implies $n=\frac53$. However some additional work is required to show that there is no rational $n$ in the range $(0,1)$ that would satisfy the problem.
In fact, the condition $n\in\mathbb{Q}$ appears to be excessive, since quite obviously $S(n)\le S(0)=\frac16$ for $0\le n\le1$. Generally, it can be shown that the problem has a unique solution $n$ for any $S(n)\in[\frac1{10},\frac12)\cup(\frac12,+\infty)$, whereas there is no solution for any other value.
A reasonable variation of the problem would be asking the same question for $S(n)=\frac18$ (or any other rational value between $\frac1{10}$ and $\frac16$). The correct answer would be that there exists no rational $n$ satisfying the problem, which can be shown by applying Lindemann–Weierstrass theorem.