Finiteness of the sums of reciprocals of positive solutions of $\tan x = x$ and $x = \tan \sqrt x$ Let $a_n$ be the sequence of positive solutions of the equation $\tan x=x$ and $b_n$ be a sequence of positive solutions of the equation $x=\tan \sqrt x$.

Prove that $\sum \dfrac{1}{a_n}$ diverges but $\sum \dfrac{1}{b_n}$ converges.

We have $\tan {a_n}=a_n$ and $b_n=\tan \sqrt b_n$.In order to check convergence of $\sum \dfrac{1}{a_n}$ and $\sum \dfrac{1}{b_n}$ we have to check convergence of $\sum \dfrac{1}{\tan a_n}$ and $\sum \dfrac{1}{\tan \sqrt b_n}$.
I am facing problem on how to judge whether the $n^{th}$ term tends to zero or not.Any help will be appreciated.
 A: For each integer $n$, $\tan x$ is a bijection from $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$ to $\mathbb{R}$. 
So for each positive integer $n$, there is exactly one positive solution to $x = \tan x$ in each interval $(\tfrac{2n-1}{2}\pi,\tfrac{2n+1}{2}\pi)$. Also, the only solution in $(-\tfrac{1}{2}\pi,\tfrac{1}{2}\pi)$ is $x = 0$, which isn't positive.
Hence, the $n$-th smallest positive solution to $x = \tan x$ satisfies $\tfrac{2n-1}{2}\pi < a_n < \tfrac{2n+1}{2}\pi$. 
Using similar logic, you can show that the $n$-th smallest positive solution to $x = \tan \sqrt{x}$ satisfies $\left(\tfrac{2n-1}{2}\pi\right)^2 < b_n < \left(\tfrac{2n+1}{2}\pi\right)^2$. 
Can you determine the convergence of $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{a_n}$ and $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{b_n}$ from this?
A: Observe that if $\tan(x) = x$ then $x \in (n\pi, n\pi+\frac{\pi}{2})$. This says $$n\pi < x < n\pi + \frac{\pi}{2} \implies \frac{1}{n\pi} > \frac{1}{x}>\frac{1}{n\pi+\frac{\pi}{2}} $$ This says that $$\sum_{n=1}^{\infty}\frac{1}{a_{n}} > \sum_{n=1}^{\infty}\frac{1}{n\pi+\frac{\pi}{2}}$$ which proves divergence of $(a_{n})$
