3
$\begingroup$

Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

Consider a sequence $x_n, n\ge1$ formed by positive solutions to $x \sin{x}=1$.

How can we find

$$\lim _{n\rightarrow \infty}(n(x_{2n+1}-2\pi n))= L$$

and

$$\lim _{n\rightarrow \infty}(n^3(x_{2n+1}-2\pi n- \frac{L}{n}))= L_2$$

?

  • 1
    You mean $2x_{n+1}$ instead of $2_{n+1}$?2012-10-11
  • 1
    I think he means $x_{2n+1}$ where he writes "2_{n+1}", otherwise the limit won't exists, I think.2012-10-11
  • 0
    @Peter Fixed. It was a typo.2012-10-11
  • 0
    I wonder if there is an elementary approach for it ...2012-10-25

4 Answers 4