let sequence $\{a_{n}\}$,such $a_{1}=2\pi-6$, and $$a_{n}=\left\lceil\dfrac{2\pi}{a_{n-1}}\right\rceil\cdot a_{n-1}-2\pi$$

Find the $$\lim_{n\to\infty}a_{n}$$

where $\left\lceil\dfrac{2\pi}{a_{n-1}}\right\rceil$is the smallest integer not less than $\dfrac{2\pi}{a_{n-1}}$

My try: since $$a_{n}>0,\dfrac{a_{n}}{a_{n-1}}=\lceil\dfrac{2\pi}{a_{n-1}}\rceil-\dfrac{2\pi}{a_{n-1}}\le 1$$ and then I can't Find this limit,and maybe other nice methods,

Thank you very much

  • $\begingroup$ Numerical calculation plus intuition yields the limit 0, but I do not have a rigorous proof yet. $\endgroup$ – Peter Jan 2 '14 at 15:32
  • 1
    $\begingroup$ The intuition : If a(n-1) is very small, the paranthese can be omitted without making a big mistake, and a(n) gets 0, but this is not a proof, of course. $\endgroup$ – Peter Jan 2 '14 at 15:34
  • $\begingroup$ wolframalpha.com/input/?i=a%5Bn%5D+%3D+RoundUp%5B2*pi%2Fa%5Bn-1%5D%5D*a_%5Bn-1%5D-2*pi%2C+a%5B1%5D+%3D+2*pi-6 $\endgroup$ – gt6989b Jan 2 '14 at 15:36
  • $\begingroup$ Since the ceiling of $2\pi/a_{n-1}$ exceeds it, the product of that ceiling by $a_{n-1}$ is at least $2\pi$, forcing $a_n>0$ provided $a_{n-1}>0.$ $\endgroup$ – coffeemath Jan 2 '14 at 16:17

$a_1 > 0$ and $\bigg\lceil\frac{2\pi}{a_n}\bigg\rceil \ge \frac{2\pi}{a_n}$. By induction you can easily prove that $a_n >0 \, \forall \, n \in \mathbb N$.

As you proved, if $a_n >0$, then the sequence is strictly decreasing.

Remains to show that the actual limit is $0$, a reasonable guess.

Using the fact that the sequence is strictly decreasing and strictly positive, you know that the sequence has a unique limit.

The limit will solve the equation $a = \big\lceil\frac{2\pi}{a}\big\rceil a - 2\pi$ which can be rewritten as : $$\bigg\lceil\frac{2\pi}{a}\bigg\rceil - \frac{2\pi}{a} = 1$$

This has no solution for any $a_1 > a > 0$ therefore the limit is not strictly positive.

So the limit is $0$.

Perhaps easier (just thought about it), $$\lim_{a_n\to0} \bigg\lceil\frac{2\pi}{a_n}\bigg\rceil a_n - 2\pi= 0$$ therefore $0$ is a limit. Since the limit is unique, it is $\underline{\text{the}}$ limit.

Another way I thought of doing this is to use the squeeze theorem. I'm sure there's a way to do it.


If there is a (nonzero) limit, it should satisfy $$a = \lceil\frac{2\pi}{a}\rceil a - 2\pi,$$ which should have a solution (which I am having some trouble finding).

  • $\begingroup$ It should satisfy $a = \left\lceil\frac{2\pi}{a}\right\rceil a - 2\pi$. $\endgroup$ – Constructor Jan 2 '14 at 16:10
  • $\begingroup$ @Constructor: ooops, reading comprehension issues. $\endgroup$ – Igor Rivin Jan 2 '14 at 16:15

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