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.
Best Answer
$-2\pi$ 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 $b_n \to 0.$