Convergence of series of solutions to $\tan(x)=x$ and $\tan(\sqrt(x)) = x$

From my solution here, you have that $p_n \sim \left(n \pi + \dfrac{\pi}2\right)$. Similarly, you can prove that $q_n \sim Cn^2$. Now by limit comparison test, you should be able to conclude what you want.

Hint: On $\left[n\pi,n\pi+\frac12\right)$, $\tan(x)$ increases monotonically from $0$ to $\infty$.

Thus, we have a root of $\tan(x)=x$ between $n\pi$ and $(n+1/2)\pi$ and a root of $\tan(\sqrt{x})=x$ between $(n\pi)^2$ and $((n+1/2)\pi)^2$.

Therefore, $\frac1{p_n}$ is between $\frac1{n\pi}$ and $\frac1{(n+1/2)\pi}$

Furthermore, $\frac1{q_n}$ is between $\frac1{(n\pi)^2}$ and $\frac1{((n+1/2)\pi)^2}$

Draw a quick picture of $y=\tan x$ and $y=x$ and identify $p_n$ in the picture. In what intervals are the $p_n$ found?

For the $q_n$, it may be easier to note that $\sqrt{q_n}$ are solutions to $\tan x=x^2$. Repeat the procedure.