How to calculate the series $ \sum_{n=1}^{+\infty}\frac{1}{x_n^2} $ if $ x_1, x_2, ..., x_n, ... $ are all positive roots for $ 2020\tan x=2021x $. 
If $ x_1, x_2, ..., x_n, ... $ are all positive roots for $ 2020\tan x=2021x $, it is easy to verify that $ \sum_{n=1}^{+\infty}\frac{1}{x_n^2} $ is convergent. However I can not calculate the exact value for the limit.

What I can do is to estimate the order of $ \frac{1}{x_n^2} $, which is not useful for the calculation. I know that the equation $ 2020\tan x=2021x $ may have some relations with the problem of eigenvalues for the wave equation, but I cannot use it to solve this problem either. Can you give me some hints or references?
 A: If you are looking for the solutions of
$$\tan(x)=k x$$ with $k>1$, beside the very first one, they are given by
$$x_n=q-\frac 1{k q}-\frac{2 k-1}{2 k^3 q^3}+\cdots \qquad \text{with} \qquad q=(2n+1)\frac \pi 2$$
Just to check with $\frac{2021}{2020}$
$$\left(
\begin{array}{ccc}
n & \text{estimate} & \text{solution} \\
 2 & 4.49551 &  4.49352 \\
 3 & 7.72569 &  7.72532 \\
 4 & 10.9043 &  10.9042 \\
 5 & 14.0663 &  14.0662 \\
 6 & 17.2208 &  17.2208 
\end{array}
\right)$$
which makes
$$\frac 1 {x_n^2}=\frac 1 {q^2}+\frac{2}{k q^4}+\cdots$$
This would give
$$\sum_{n=2}^\infty\frac 1 {x_n^2}=\frac 12+\frac 1 k\left(\frac{1}{3}-\frac{32}{\pi ^4}\right)-\frac{4}{\pi ^2}$$
Concerning the first root, using Taylor series
$$\tan(x)-kx=(1-k) x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ which gives
$$x_1^2=\frac{1}{4} \left( \sqrt{5(24 k-19)}-5\right)\implies \frac 1{x_1^2}=\frac 4 {\sqrt{5(24 k-19)}-5}$$
For your value of $k$, this gives $x_1=0.0385261897$ while the exact solution is                                     $x_1=0.0385261829$
$$\sum_{n=1}^\infty\frac 1 {x_n^2}=\frac 4 {\sqrt{5(24 k-19)}-5}+\frac 12+\frac 1 k\left(\frac{1}{3}-\frac{32}{\pi ^4}\right)-\frac{4}{\pi ^2}$$
If you perform the summation from $n=1$, this leads to a large number $(673.832630772193)$ which is not recognized by inverse symbolic calculators.
