Find $\,\lim\limits_{n\to \infty}a_n\,$ such that $\,a_n=\left\lceil{\frac{2\pi}{a_{n-1}}}\right\rceil\!\cdot\!a_{n-1}-2\pi$ . 
Question : Let $\{a_n\}$ be a real sequence such that $\,a_1=2\pi-6\;$ and $\;a_n=\left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil\!\cdot\!a_{n-1}-2\pi\;,$
where $\left\lceil\cdot\right\rceil\;$ is the smallest integer not less than $\dfrac{2\pi}{a_{n-1}}$.
Then $\;\lim\limits_{n\to\infty}a_n=\ldots$

$\{a_n\} $ is a monotonically decreasing sequence.
I presumed that if $\,\lim\limits_{n\to \infty} a_n=l\,,\,$ i.e. convergent, then $l=\left\lceil\dfrac{2\pi}{l}\right\rceil\!\cdot\!l-2\pi\implies l\left(1-\left\lceil\dfrac{2\pi}{l}\right\rceil\right)=-2\pi$.
Afterwards I was looking for the possible solutions and found $l=-2\pi^+,\,$ i.e. $\,-2\pi\,$ from the right hand.
I think this is not enough to say that the limit is $\,-2\pi^+.$
I am looking for a better solution to this as I have used a graphing calculator for the same, so my reasoning is incomplete. Thanks.
 A: $-2\pi$ is a rather off conclusion, it might be that you are confusing the ceiling and floor functions. The limit in fact is $0$.
Note that the ceiling function is larger than its argument (the smallest integer not less than). So, if $a_{n-1} > 0,$ then $$a_n = \lceil 2\pi /a_{n-1}\rceil a_{n-1} - 2\pi \ge 2\pi/a_{n-1} \cdot a_{n-1} - 2\pi = 0.$$ Since $a_1 > 0,$ we inductively conclude that the sequence is non-negative. Further, since $\lceil x \rceil < x + 1,$ (why?), $$ a_n < (2\pi/a_{n-1} + 1) a_{n-1} - 2\pi = a_{n-1},$$ and so we have a bounded decreasing sequence, for which a limit must exist (Bolzano-Weierstrass). In particular, note that $a_n$ must approach this limit from above.
Let us define $L := \lim a_n$. From the above, we know that $L\ge 0.$ We claim that $L = 0,$ which we shall prove by contradiction.
Indeed, suppose that $L > 0.$ Then notice that $\lim \frac{2\pi}{a_n} = 2\pi/L$ and $\lim a_n/a_{n-1} = 1.$ Further, for any $n, a_n > L$. We can rewrite our iteration as $$ \frac{a_{n+1}}{a_{n}} = \lceil 2\pi/a_n\rceil - 2\pi/a_n,$$ and so conclude that $$ \lim \lceil 2\pi/a_n\rceil = 2\pi/L + 1.$$
There are two cases, that $2\pi/L$ could be an integer or a non-integer. Firstly, if $2\pi/L$ is not an integer, then notice that it is a continuity point of $\lceil x \rceil,$ so we have $$ \lim \lceil 2\pi/a_n \rceil = \lceil 2\pi/L\rceil,$$ but this contradicts $\lim \lceil 2\pi/a_n\rceil = 2\pi/L + 1$ since $\lceil x \rceil < x + 1$ for any real $x$. So if $L > 0, 2\pi/L$  must be an integer.
But notice that $2\pi/a_n < 2\pi/L$, since $a_n > L$. So if $2\pi/L$ is an integer, then we then we would instead have $$ \lim \lceil 2\pi/a_n\rceil \le \lim \lceil 2\pi/L\rceil = 2\pi/L < 2\pi/L + 1,$$ again a contradiction. We conclude that $2\pi/L$ cannot be an integer either.
Since $2\pi/L$ is neither a non-integer nor an integer, it is not a positive real number. Therefore $L \le 0,$ and applying the non-negativity of $L$ leaves us with the conclusion $L = 0.$

Since your question suggests that you were confusing the ceiling and floor functions, we can also look at the floor case. This uses the same idea, although it is slightly more complicated.
Let $b_1 = (2\pi -6),$ and $$b_n := \lfloor \frac{2\pi}{b_{n-1}}\rfloor b_{n-1} - 2\pi.$$
By direct computation, $b_2 < 0,$ so we must account for both positive and negative $b_n$. Since $$ (2\pi/b_{n-1} - 1) < \lfloor 2\pi/b_{n-1} \rfloor \le 2\pi/b_{n-1},$$ we can conclude hta $$b_{n-1} < 0 \implies -b_{n-1} > b_n \ge 0 \\ b_{n-1} \ge 0 \implies -b_{n-1} < b_n \le 0. $$
Notice that this means that $|b_n|$ is strictly decreasing, and so at least this must have a limit. Of course, if $\lim |b_n| = 0,$ then $b_n$ itself must have the limit $0$ too. Let $c_n = (-1)^{n+1} b_n = |b_n|$. We will indeed argue that $c_n \to 0.$
Let $L:= \lim c_n,$ and for contradiction assume it is $> 0.$ Note that this means $c_n > 0$ for all $n$. Using the same idea as above, diving throughout by $b_{n-1},$ and using the notation $\{x\} = x - \lfloor x \rfloor,$ we have that $$ \frac{b_{n+1}}{b_{n}}  = \left \lfloor \frac{2\pi}{b_n} \right\rfloor - \frac{2\pi}{b_{n}} = - \{ 2\pi/b_n\} \\ \iff \frac{c_{n+1}}{c_n} = \{ (-1)^{n+1} 2\pi/c_n\}.$$
Observe that for any $x \not \in \mathbb{Z},$ $\{-x\} = 1-\{x\},$  while for $x \in \mathbb{Z}, \{-x\} = \{x\} = 0.$ Note also that if $2\pi/c_n$ were ever an integer, then this would force $c_{n+1} = 0,$ and since $c_n \ge 0,$ would already force $L = 0.$ So we may assume that $2\pi/c_n$ is never an integer. Then we have that \begin{align} n \textrm{ even} \implies c_{n+1}/c_n &= 1- \{2\pi/c_n\} \\ n \textrm{ odd} \implies c_{n+1}/c_n &= \{2\pi/c_n\}.\end{align}
But we know that $c_{n+1}/c_n \to 1,$ so it has to be the case that $$ \{2\pi/c_{2n}\} \to 0, \{2\pi/c_{2n+1}\} \to 1.$$ Now, if $L$ was such that $2\pi/L$ is a non-integer, then the above can't happen because $2\pi/c_n \to 2\pi/L$ and this limit is a continuity point of $\{x\}$, and so the limit point of the even index subsequence and the odd index subsequence must agree. So it must be the case that $2\pi/L$ is an integer. But $c_n \searrow L,$ so for large $n$, $2\pi/c_n$ is slightly smaller than an integer, and therefore $\{2\pi/c_n\} \to 1,$ which contradicts the behaviour for even $n$. We conclude that $L > 0$ cannot hold.
Therefore, $c_n = |b_n| \to 0,$ and thus $b_n \to 0.$
A: Recall the following lemmata:
$A)$ If a sequence $(x_n)\subset \mathbb R$ converges, then it converges to a unique limit.
$B)$ If a sequence $(x_n)\subset \mathbb R$ is monotonic and bounded, then the sequence $(x_n)$ converges.
$C)$ If a sequence $(x_n)\subset \mathbb R$ is convergent such that for some $a,b\in \mathbb R, a\le x_n\le b$ for all $n$, then $a\le \lim x_n\le b$.
Claim: In order to prove that $\lim a_n=0$, it suffices to prove $(1)$ and $(2)$ below:
$1)$ $a_n\gt 0$ for all $n\in \mathbb N$.
$2)$ $a_n$ is monotonically decreasing.
Proof: Suppose that $1)$ and $2)$ are true. Then $a_n$ is bounded (above by $a_1$ and below by $0$) so by $(B)$, $(a_n)$ converges, and by $(A) \lim a_n=p$, say. It follows by $(C)$ and $(1)$ that $p\ge 0$.
Suppose on the contrary that $p>0$. It follows that $\frac{2\pi}{a_n}$ monotonically increases to $\frac{2\pi}p$. For all large $n$, $\frac{2\pi}{a_n}$ is very close to $\frac{2\pi}p$, whence it follows that $\lceil \frac{2\pi}{a_n}\rceil=\lceil \frac{2\pi}p\rceil$ for all large $n$ (For more details on this, please see the last section of this post). It follows that $\lim_n\lceil \frac{2\pi}{a_n}\rceil=\lceil \frac{2\pi}p\rceil$.
From the given equation, we have $\lceil \frac{2\pi}p\rceil.p-2\pi=p,$ whence $p=\lceil \frac{2\pi}p\rceil.p-2\pi\lt (\frac{2\pi}p+1)p-2\pi=p\implies p<p$, which is a contradiction. It follows that $p=0$. This completes the proof.
So if we prove $(1)$ and $(2)$, then we are done. But $(1)$ and $(2)$ follow by induction so we are done.

Response to comment:
Lemma:
If $(x_n)\subset \mathbb R$ is a monotonically increasing bounded sequence such that $x_n\to x$, then $\lceil x_n\rceil \to \lceil x \rceil$.
Proof: Case $1$: $x$ is an integer.
There exists $N$ such that $n\ge N\implies x-\frac 12 <x_n\le x_{n+1}\le x$. If $x$ is an integer, then indeed $\lceil x_n\rceil =\lceil x\rceil$ for all $n\ge N$.
Case $2$: $x$ is not an integer.
In this case, let $d=\frac 12\min(x-\lfloor x\rfloor, \lceil x\rceil-x)$.
There exists $N'$ such that $n\ge N'\implies -d<x_n-x<d\implies x-d<x_n\le x\implies x-d<x_n\le x_{n+1}\le x\;,\;$
whence it follows that
$\lceil x_n\rceil = \lceil x\rceil$. QED.
A: From the recursion equation, we have
$$\;a_n=\left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil\!\cdot\!a_{n-1}-2\pi$$
$$\Rightarrow \; \dfrac{a_n}{a_{n-1}}=\left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil - \dfrac{2\pi}{a_{n-1}} = \left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil - \left(\left\lfloor{\dfrac{2\pi}{a_{n-1}}}\right\rfloor + \left\{\dfrac{2\pi}{a_{n-1}}\right\}\right)$$
$$\Rightarrow \; \dfrac{a_n}{a_{n-1}} = \left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil - \left(\left\lceil{\dfrac{2\pi}{a_{n-1}}}\right\rceil - 1 + \left\{\dfrac{2\pi}{a_{n-1}}\right\}\right)$$
$$\Rightarrow \; \dfrac{a_n}{a_{n-1}} = 1 - \left\{\dfrac{2\pi}{a_{n-1}}\right\}$$
where $\left\lfloor  \right\rfloor$ is floor function and $\left\{\right\}$ is the fractional part function. Also, seeing the confusion in nomenclature in the comments, I reiterate that $x = \left\lfloor  x \right\rfloor + \left\{x\right\}$.
So,
$$ \lim_\limits{n\to\infty}\dfrac{a_n}{a_{n-1}} = \lim_\limits{n\to\infty} \left(1 - \left\{\dfrac{2\pi}{a_{n-1}}\right\}\right)$$
Assuming that $\lim_\limits{n\to\infty} a_n$ exists finitely and non-zero, we can write
$$ \dfrac{\lim_\limits{n\to\infty} a_n}{\lim_\limits{n\to\infty} a_{n-1}} = 1 - \left\{\dfrac{2\pi}{\lim_\limits{n\to\infty}  a_{n-1}}\right\}$$
$$ \Rightarrow  1 = 1 - \left\{\dfrac{2\pi}{\lim_\limits{n\to\infty}  a_{n}}\right\}$$
$$ \Rightarrow  \left\{\dfrac{2\pi}{\lim_\limits{n\to\infty}  a_{n}}\right\} = 0$$
This implies that $\lim_\limits{n\to\infty}  a_{n} = \infty$ which contradicts the assumption made.
Hence, this concludes that there can be 2 possibilities:

*

*$\lim_\limits{n\to\infty}  a_{n} = 0$.

OR


*$\lim_\limits{n\to\infty}  a_{n} = \pm\infty$.

Now, from here, we can narrow down the answer as follows:
For some arbitrary N, we have
$$\prod_{n=2}^{N} \dfrac{a_n}{a_{n-1}} = \prod_{n=2}^{N} \left(1 - \left\{\dfrac{2\pi}{a_{n-1}}\right\}\right)$$
$$\Rightarrow \dfrac{a_N}{a_{1}} = \left(1 - \left\{\dfrac{2\pi}{a_{1}}\right\}\right)\left(1 - \left\{\dfrac{2\pi}{a_{2}}\right\}\right)\left(1 - \left\{\dfrac{2\pi}{a_{3}}\right\}\right) \ldots \left(1 - \left\{\dfrac{2\pi}{a_{N-2}}\right\}\right)\left(1 - \left\{\dfrac{2\pi}{a_{N-1}}\right\}\right) \le 1$$
$$\Rightarrow \dfrac{a_N}{a_{1}} \le 1$$
$$\Rightarrow \lim_\limits{N\to\infty} a_N \le \lim_\limits{N\to\infty} a_{1} = 2\pi - 6$$
And it can be also shown that $a_N$ is always positive.
This points to the fact that
$$\boxed{\lim_\limits{n\to\infty}  a_{n} = 0}$$
Comment: However, if the ceiling functions in the above question are actually floor functions in which case, the limit exists.
