How find this $\lim_{n\to\infty}a_{n}$ 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
$$a_{n}&gt;0,\dfrac{a_{n}}{a_{n-1}}=\lceil\dfrac{2\pi}{a_{n-1}}\rceil-\dfrac{2\pi}{a_{n-1}}\le 1$$
and then I can't Find this limit,and maybe other nice methods,

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

As you proved, if $a_n &gt;0$, 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 $a_1 &gt; a &gt; 0$ 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).