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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.

TheStudent
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learner
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  • 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

3 Answers3

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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.

Community
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  • @learner Compare $1/p_n$ with $1/l_n$ and note that $\dfrac{1/p_n}{1/l_n} \sim 1$. –  May 10 '13 at 17:56
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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 $\frac1{((n+1/2)\pi)^2}$

Yadati Kiran
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robjohn
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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.

Harald Hanche-Olsen
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