-2

Let $\{a_n\}$ be the sequence of consecutive positive solutions of the equation $\tan x=x$ and $\{b_n\}$ be the sequence of consecutive positive solutions of the equation $\tan \sqrt x=x$. Then test the convergence of the series $\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$ and $\sum\limits_{n=1}^{\infty}\frac{1}{b_n}$.

From given conditions we have $\tan a_n=a_n$ and $\tan \sqrt{b_n}=b_n$. Then , $a_n=\tan^{-1}a_n$ and then $\tan a_n=\tan^{-1}a_n$. But from here how I can test the convergence of the series $\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$ ?

If I can show where the solutions of $x=\tan x$ belong then I can do it self.

$x=0$ is a solution of the equation $x=\tan x$. In the interval $(0,\pi/2)$ , $x<\tan x$. But in which interval the positive solutions belong ?

I saw this answer , but here it is stated as a Theorem , but it contains no proof..

Can anyone help me??

amWhy
  • 209,954
Empty
  • 13,012
  • 1
    The second paragraph (supposedly indicating what the OP tried) reads phony at best. To at least start from something, what can you say about the location of $a_n$? – Did Oct 19 '15 at 12:21
  • the $a_n$ are near the $x$ coordinates of the $y$ asymptotes, so you can use the comparison test on $b_n$, where $b_n$ is the $x$ coordinate of the $n$th asymptote – Zarrax Oct 19 '15 at 16:49
  • @Did ) I can't understand the reason of downvoting and close vote.. – Empty Oct 19 '15 at 17:31
  • @S.Panja-1729 To better your understanding of the site, you might want to read and ponder some howtoask page. After more than 1 year and 150 questions on the site, avoiding to post duplicates (since, unless you explain what you are missing on the other page, which answers every explicit question you ask here, say "in which interval the positive solutions belong ?" (!!), this question is truly a duplicate) would be a start. Oh, and "it contains no proof" is simply not true. – Did Oct 19 '15 at 19:08
  • 4
    Duplicate. – metamorphy Dec 19 '22 at 05:02

1 Answers1

2

Looking at the graphs of the functions $x\mapsto x$ and $x\mapsto\tan x$ for $x>0$ we see that the $n$th positive solution $a_n$ of the equation $\tan x=x$ satisfies $$n\pi<a_n<\left(n+{1\over2}\right)\pi\ .$$ This immediately implies that $\sum_{n=1}^\infty{1\over a_n}$ diverges.

The $b_n$ however are in bijective correspondence with their square roots $c_n:=\sqrt{b_n}$, and the $c_n>0$ satisfy the equation $\tan x=x^2$. Looking at the graphs of $x\to x^2$ and $x\to\tan x$ for $x>0$ we see that in each interval $\ \bigl]n\pi,\bigl(n+{1\over2}\bigr)\pi\bigr[$ there is exactly one solution $c_n$ of this equation, so that $$n\pi<c_n<\left(n+{1\over2}\right)\pi\ .$$ From this we conclude that $${1\over b_n}={1\over c_n^2}<{1\over \pi^2 n^2}\ ,$$ which then implies that $\sum_{n=1}^\infty{1\over b_n}$ converges.

Christian Blatter
  • 226,825