$x_n$ is the sequence of all the positive roots of the equation $tan (x)=x$. Is the series $∑1/{x^2_n}$ convergent? 
let ${x_n}$ be the sequence of the positive roots of the equation $tan (x)=x$
Prove/disprove: the series $∑1/{x^2_n}$ converges.

I spent a lot of time trying to prove it, but I do not think I'm doing well.
By looking at the graph, I know that for any natural n, $nπ<x_n<(n+1/2)π$
So let $a_n=1/((n+1/2)π)^2$ and $c_n=1/(nπ)^2$$
Using the comparison test with the convergent series $ ∑1/{n^2}$ and $p_n = 1/{n^2}$ we have:
$\lim_{n\rightarrow∞} (a_n/p_n) = \lim_{n\rightarrow∞} (1/((n+1/2)π)^2 )/(1/n^2 ) = 1/\pi^2$
$\lim_{n\rightarrow∞} (c_n/p_n) = \lim_{n\rightarrow∞} (1/(nπ)^2 )/(1/n^2 ) = 1/\pi^2$
So,  $ ∑1/{a_n}$ and $ ∑1/{c_n}$ are convergent.
let $ε>0$.
There is some N such that for any $n,k>N$ , $ |∑1/{a_n}| < ε$ and $ |∑1/{c_n}|< ε$
We know that $a_n≤1/(x_n^2 )≤c_n$, so $ ∑1/{a_n}≤∑1/{x^2_n}≤∑1/{c_n}$
Therefore $-ε<∑1/{x^2_n}<ε$
Which leading the conclusion: according to Cauchy convergence test, the series $∑1/{x^2_n}$ convergent.
Questions:
The graph conclusions are correct?
How can I prove mathematically what I assumed from the graph?
Is the rest of the proof correct?
Can someone prove or disprove the problem from zero in a formal way? I have been reading every bit of information on Stack and the Internet, I couldnt find a full proof to this problem.
 A: As to the zeros of $\tan x$, we know that $\lim_{x\to n\pi+}\tan x = -\infty$, and $\lim_{x\to n(\pi+1/2)-}\tan x = \infty$ and that $\tan x$ is increasing, so there is exactly one zero between $n\pi$ and $(n+1/2)\pi.$
Your proof is correct, I think, but it can be simplified.  Since $x_n>n\pi$, we have ${1\over x_n^2}<{\pi\over n^2}$, so the series converges be comaparison with $\sum{\pi\over n^2}$. 
A: You have $$ x_n>n\pi \implies \frac {1}{x_n^2}<\frac {1}{n^2}$$
Thus the series converges by comparison.
A: To prove that in each interval $(n\pi-\frac\pi{2}, n\pi+\frac\pi{2})$, there exists only one solution of equation $\tan x = x$ note that on each integral $\tan x$ is continuous and
$$\frac{d}{dx}(\tan x - x) = \tan^2 x \ge 0$$
If in such interval there were two numbers $x_1$, $x_2$, $x_1<x_2$ such that $\tan x_1 - x_1 = \tan x_2 -x_2 =0$ it would mean that 
$$ 0 = \int_{x_1}^{x_2} \frac{d}{dx}(\tan x - x) \, dx= \int_{x_1}^{x_2} \tan^2 x \, dx> 0$$
which is a contradiction.
This proves that in each interval there exists $(n\pi-\frac\pi{2}, n\pi+\frac\pi{2})$  at most one solution. To prove that there exists exactly one solution, it is enough to note that 
$$\lim_{x\rightarrow (n\pi-\frac\pi{2})^+} (\tan x -x) = -\infty , \qquad \lim_{x\rightarrow (n\pi+\frac\pi{2})^-} (\tan x -x) = \infty$$
and use the continuity of $\tan x-x$ on the interval $(n\pi-\frac\pi{2}, n\pi+\frac\pi{2})$. Using similar argument you can prove that for $n>0$ this unique solution lies in the interval $(n\pi, n\pi+\frac\pi{2})$
The rest of the proof is correct, although you can make it simpler, it is enough to note that $\frac{1}{x_n^2} < \frac{1}{\pi^2n^2}$ and series $\sum_n \frac{1}{n^2}$ is convergent.
