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

I am stuck on the following problem:

Let $\{p_n\}$ be the sequence of consecutive positive solutions of the equation $\tan x=x$ and let $\{q_n\}$ be the sequence of consecutive positive solutions of the equation $\tan \sqrt x=x.$ Then how can I prove that $\sum_{n=1}^{\infty} \frac{1}{p_n}$ diverges but $\sum_{n=1}^{\infty} \frac{1}{q_n}$ converges ?

Can someone point me in the right direction?Thanks in advance for your time.

• For the first one, it gets close to $p_n \approx n$ (you can probably cook up a crude estimate that is precise enough to bound the sum from below by e.g. half the harmonic series). – vonbrand May 10 '13 at 17:23

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.

• @learner Compare $1/p_n$ with $1/l_n$ and note that $\dfrac{1/p_n}{1/l_n} \sim 1$. – user17762 May 10 '13 at 17:56

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.