Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\sin x$

Let tangents be drawn to the curve $y=\sin x$ from the origin. Let the points of contact of these tangents with the curve be $(x_k,y_k)$ where $x_k\gt 0; k\ge 1$ such that $x_k\in (\pi k, (k+1)\pi)$ and $$a_k=\sqrt {x_k^2+y_k^2}$$ (Which is basically the distance between the corresponding point of contact and the origin i.e. the length of tangent from origin) .

I wanted to know the value of

$$\sum_{k=1}^{\infty} \frac {1}{a_k ^2}$$

Now this question has just popped out in my brain and is not copied from any assignment or any book so I don't know whether it will finally reach a conclusion or not.

I tried writing the equation of tangent to this curve from origin and then finding the points of contact but did not get a proper result which just that the $x$ coordinates of the points of contact will be the positive solutions of the equation $\tan x=x$

On searching internet for sometime about the solutions of $\tan x=x$ I got two important properties of this equation. If $(\lambda _n)_{n\in N}$ denote the roots of this equation then

$$1)\sum_n^{\infty} \lambda _n \to \infty$$ $$2)\sum_n^{\infty} \frac {1}{\lambda _n^2} =\frac {1}{10}$$

But were not of much help.

I also tried writing the points in polar coordinates to see if that could be of some help but I still failed miserably.

I could not think of any method so any other method would be openly welcomed.

Any help would be very beneficial to solve this problem.

Thanks in advance.

Edit:

On trying a bit more using some coordinate geometry I found that the locus of the points of contact is $$x^2-y^2=x^2y^2$$

Hence for sum we just need to find $$\sum_{k=1}^{\infty} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{2\lambda _k ^2} +\sum_{k=1}^{\infty} \frac {1}{2(\lambda _k ^2 +2)} =\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2} $$

Now for the second summation I did think about it to form a series but for the roots to be $\lambda _k^2 +2$ we just need to substitute $x\to \sqrt {x−2}$ in power series of $\frac {\sin x-x\cos x}{x^3}$ and then get the result but it was still a lot confusing for me.

Using $x\to\sqrt {x-2}$ in the above power series and using Wolfy I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second summation equal to $$\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}?$$

Is this value correct or did I do it wrong?

I would also like to know if there is some other method to solve this problem

The points of contact are where the tangent to $y=\sin(x)$, which has a slope of $\cos(x)$, has the same slope as the line from the origin, $\frac{\sin(x)}x$. Thus, we are looking at the points where $x_k=\tan(x_k)$.

The square of the length of the line from the origin is $x_k^2+\sin^2(x_k)=\frac{x_k^4+2x_k^2}{x_k^2+1}$. Therefore, the sum we are looking for is $$ \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2}\tag1 $$

The residue of $f(z)=\frac1{\tan(z)-z}-\frac1{(z^2+2)(\tan(z)-z)}$ where $z\ne0$ and $\tan(z)=z$ is $$ \frac1{z^2}-\frac1{z^4+2z^2}=\frac{z^2+1}{z^4+2z^2}\tag2 $$

Thus, the sum of all the residues of $f(z)$ is $2$ times the sum we are seeking plus the residue of $f(z)$ at $z=0$, which is $\frac3{20}$, and the sum of the residues of $f(z)$ at $z=\pm i\sqrt2$, which is $-\frac1{2-\sqrt2\tanh(\sqrt2)}$

Note that the limit $$ \lim_{k\to\infty}\int_{\gamma_{k,\lambda}}f(z)\,\mathrm{d}z=\int_{\gamma_\lambda}f(z)\,\mathrm{d}z\tag3 $$ where $k\in\mathbb{Z}$ and the paths are $$ \scriptsize\gamma_{k,\lambda}=[k\pi+i\lambda,-k\pi+i\lambda]\cup\underbrace{[-k\pi+i\lambda,-k\pi-i\lambda]}_{\le\frac{2\lambda}{k\pi}}\cup[-k\pi-i\lambda,k\pi-i\lambda]\cup\underbrace{[k\pi-i\lambda,k\pi+i\lambda]}_{\le\frac{2\lambda}{k\pi}}\tag4 $$ and $$ \gamma_\lambda=(\infty+i\lambda,-\infty+i\lambda)\cup(-\infty-i\lambda,\infty-i\lambda)\tag5 $$ and $2\pi i$ times the sum of all the residues of $f(z)$ is $$ \lim_{\lambda\to\infty}\int_{\gamma_\lambda}f(z)\,\mathrm{d}z=-2\pi i\tag6 $$

$(6)$ means the sum of the residues of $f(z)$ over all singularities is $-1$. This is $2$ times the sum we are looking for plus $\frac3{20}-\frac1{2-\sqrt2\tanh(\sqrt2)}$

Therefore, $$ \bbox[5px,border:2px solid #C0A000]{ \begin{align} \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2} &=-\frac{23}{40}+\frac1{4-2\sqrt2\tanh\left(\sqrt2\right)}\\ &=0.097374597898595746715 \end{align} }\tag5 $$

Numerical Check

Note that each of the roots is a little less than an odd multiple of $\frac\pi2$:

$x_1=4.4934094579090641753\approx\frac{3\pi}2$
$x_2=7.7252518369377071642\approx\frac{5\pi}2$
$x_3=10.904121659428899827\approx\frac{7\pi}2$
$x_4=14.066193912831473480\approx\frac{9\pi}2$

Thus, we can under-approximate the sum using $$ \begin{align} \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2} &\approx\sum_{k=1}^\infty\frac{(2k+1)^2\pi^2/4+1}{(2k+1)^4\pi^4/16+(2k+1)^2\pi^2/2}\\ &=0.092481600740508343614 \end{align} $$

Answer to original question

Simple bound $\pi k\leq a_k \leq \sqrt{\pi^2(k+\frac12)^2+1}$ shows that $\dfrac{a_k}{a_{k+1}}\to 1$. So both sums diverge.

Answer to modified question

Again, (1) diverges. (2) also diverges since squaring the ratio doesn't change $\to 1$. (3) converges since you have $\pi k<\lambda_k=x_k<a_k$ giving $\dfrac{1}{a_k^2}\leq\dfrac{1}{\lambda_k^2}\leq\dfrac{1}{k^2}$. This is of course a very loose bound.

Finding $\displaystyle\sum_{k=1}^\infty\frac{1}{\lambda_k^2+2}$ for use in $\sum a_k^{-2}$

Recall one way of finding $\displaystyle\sum_{k=1}^\infty\lambda_k^{-2}=\frac{1}{10}$ is write down the series expansion of $$ \sin x-x\cos x $$ and set that to zero, reading off the lowest terms $$ x^3\left(\frac{1}{3}-\frac{x^2}{30}+\frac{x^4}{840}+\dots\right)=0 $$ and after cancelling $x^3$ factor in front, you read off $\dfrac{1/30}{1/3}$ in a reciprocal Viete's formula way (except to make it rigourous you need to do it properly with infinite products, but that's another story).

So now we want to do this with $\lambda_k^2+2$. You want to construct a series whose roots are $\lambda_k^2+2$ The simplest heuristic way is to use the full series expansion above (ignoring the $x^3$) and try to express it as a power series in $x^2+2$, and read off the sum of reciprocals of roots.