Define $S(n) = A(1) + A(2) + \dots + A(n)$. For each permutation $S$, we separate it into two sequences of clockwise consecutive integers $S_1$ and $S_2$, whose starting points are $n$ and $n-1$ respectively. Note that $S_1$ or $S_2$ can be one integer long. We call one of $S_1$ or $S_2$ completely reduced if all numbers other than its starting point are removed. Since the largest number is the starting point for both $S_1$ and $S_2$, the total number of descending chains in $S_1$ and $S_2$ is $G(S_1)$ and $G(S_2)$ respectively. If $S_1$ is completely reduced before $S_2$ is completely reduced, then at the time $S_1$ is completely reduced, $n-1$ is to the right of $n$, and since $S_2$ is not yet completely reduced, $n$ will just attach itself to the descending chain that includes $n-1$. Essentially, $n$ replaces $n-1$ as the starting point of $S_2$ until it is completely reduced. Thus, in this case, $G(S)=G(S_1)+G(S_2)$. If $S_2$ is completely reduced before $S_1$ is completely reduced, then at the time $S_2$ is completely reduced, $n-1$ is to the left of $n$, and since all other numbers are less than $n-1$, $n-1$ will not be a part of a descending chain until $S_1$ is completely reduced. Then only $n$ and $n-1$ remain on the circle, which form the last descending chain. Thus, in this case, $G(S)=G(S_1)+G(S_2)+1$. If $S_1$ and $S_2$ are completely reduced simultaneously, then when both of them are completely reduced, only $n$ and $n-1$ remain, which form the last descending chain. Thus, in this case, $G(S)=G(S_1)+G(S_2)+1$. For each permutation $S$, let $S’$ be the permutation formed when $n$ and $n-1$ are swapped. If $S_1$ is first to completely reduce in $S$, then $S_2$ is first to completely reduce in $S’$ and vice versa. If $S_1$ and $S_2$ completely reduce simultaneously in $S$, then they completely reduce simultaneously in $S’$ as well. For each permutation $S$, if $S_1$ and $S_2$ completely reduce simultaneously, then $\frac{G(S_1)+G(S_2)}{2}=G(S_1)+G(S_2)+1$. Otherwise, $\frac{G(S_1)+G(S_2)}{2}=G(S_1)+G(S_2)+\frac{1}{2}$. Note that $S_1$ can be any permutation of length $i$ such that $i=1,2,\dots,n-1$, and $S_2$ can be any permutation of length $n-i$. Note that $\frac{G(S_1)+G(S_2)}{2} \ge G(S_1)+G(S_2)+\frac{1}{2}$. Taking the average of both sides, we get $$A(n) \ge \frac{1}{n-1} \sum_{i=1}^{n-1} (A(i)+A(n-i)) + \frac{1}{2} = \frac{2S(n-1)}{n-1} + \frac{1}{2}.$$If $n$ and $n-1$ are next to each other, then it is impossible for $S_1$ and $S_2$ to completely reduce simultaneously. There is a $\frac{1}{n-1}$ chance that $n-1$ will be next to $n$, so there is at least a $\frac{1}{n-1}$ chance that $\frac{G(S_1)+G(S_2)}{2} \ge G(S_1)+G(S_2)+\frac{1}{2}$ and at most a $\frac{n-2}{n-1}$ chance that $\frac{G(S_1)+G(S_2)}{2} \ge G(S_1)+G(S_2)+1$. We get $$A(n) \le \frac{1}{n-1} \sum_{i=1}^{n-1} (A(i)+A(n-i)) + \frac{n-2}{n-1} + \frac{1}{2} \cdot \frac{1}{n-1} = \frac{2S(n-1)}{n-1} + \frac{2n-3}{2n-2}.$$Thus, $$\frac{2S(n-1)}{n-1} + \frac{1}{2} \le A(n) \le \frac{2S(n-1)}{n-1} + \frac{2n-3}{2n-2}.$$Adding $S(n-1)$ to all three sides, we get $$\frac{(n+1)S(n-1)}{n-1} + \frac{1}{2} \le S(n) \le \frac{(n+1)S(n-1)}{n-1} + \frac{2n-3}{2n-2},$$which can be rearranged to get $$S(n) + \frac{n}{2} \ge \frac{n+1}{n-1}\left(S(n-1) + \frac{n-1}{2}\right)$$and $$S(n)+n-\frac{1}{4} \le \frac{n+1}{n-1}\left(S(n-1)+(n-1)-\frac{1}{4}\right)$$Let $X_n = S(n) + \frac{n}{2}$ and $Y_n = S(n) + n - \frac{1}{4}$. So we have $$X_n \ge \frac{n+1}{n-1}X_{n-1} \iff \frac{X_n}{X_{n-1}} \ge \frac{n+1}{n-1} \implies \frac{X_n}{X_k} = \frac{X_n}{X_{n-1}} \cdot \frac{X_{n-1}}{X_{n-2}} \cdot \ldots \cdot \frac{X_{k+1}}{X_k}$$$$\ge \frac{n+1}{n-1} \cdot \frac{n}{n-2} \cdot \ldots \cdot \frac{k+2}{k} = \frac{n(n+1)}{k(k+1)} \iff X_n \ge \frac{n(n+1)}{k(k+1)}X_k$$$$S(n) + \frac{n}{2} \ge \frac{n(n+1)}{k(k+1)}\left(S(k) + \frac{k}{2} \right)$$for all integers $k \ge 2$. Similarly, we also have $$Y_n \le \frac{n(n+1)}{k(k+1)}Y_k$$$$S(n) + n - \frac{1}{4} \le \frac{n(n+1)}{k(k+1)}\left(S(k) + k - \frac{1}{4} \right)$$for all integers $k \ge 2$. Next, we find the value of $S(5)$. It is easy to see that $A(1) = 0, A(2) = 1, A(3) = \frac{3}{2}, A(4) = \frac{7}{3}$. To get $A(5)$, we calculate the number of descending chains in each of the $24$ permutations. Assume that $5$ is the first term in each of those permutations. $5, 4, 3, 2, 1 \to 5$ --- $1$ descending chain $5, 4, 3, 1, 2 \to 5, 2 \to 5$ --- $2$ descending chains $5, 4, 2, 3, 1 \to 5, 3 \to 5$ --- $3$ descending chains $5, 4, 2, 1, 3 \to 5, 3 \to 5$ --- $2$ descending chains $5, 4, 1, 3, 2 \to 5, 3 \to 5$ --- $3$ descending chains $5, 4, 1, 2, 3 \to 5, 2, 3 \to 5, 3 \to 5$ --- $3$ descending chains $5, 3, 4, 2, 1 \to 5, 4 \to 5$ --- $3$ descending chains $5, 3, 4, 1, 2 \to 5, 4, 2 \to 5$ --- $3$ descending chains $5, 3, 2, 4, 1 \to 5, 4 \to 5$ --- $3$ descending chains $5, 3, 2, 1, 4 \to 5, 4 \to 5$ --- $2$ descending chains $5, 3, 1, 4, 2 \to 5, 4 \to 5$ --- $3$ descending chains $5, 3, 1, 2, 4 \to 5, 2, 4 \to 5, 4 \to 5$ --- $3$ descending chains $5, 2, 4, 3, 1 \to 5, 4 \to 5$ --- $3$ descending chains $5, 2, 4, 1, 3 \to 5, 4, 3 \to 5$ --- $3$ descending chains $5, 2, 3, 4, 1 \to 5, 3, 4 \to 5, 4 \to 5$ --- $4$ descending chains $5, 2, 3, 1, 4 \to 5, 3, 4 \to 5, 4 \to 5$ --- $4$ descending chains $5, 2, 1, 4, 3 \to 5, 4 \to 5$ --- $3$ descending chains $5, 2, 1, 3, 4 \to 5, 3, 4 \to 5, 4 \to 5$ --- $3$ descending chains $5, 1, 4, 3, 2 \to 5, 4 \to 5$ --- $3$ descending chains $5, 1, 4, 2, 3 \to 5, 4, 3 \to 5$ --- $3$ descending chains $5, 1, 3, 4, 2 \to 5, 3, 4 \to 5, 4 \to 5$ --- $4$ descending chains $5, 1, 3, 2, 4 \to 5, 3, 4 \to 5, 4 \to 5$ --- $4$ descending chains $5, 1, 2, 4, 3 \to 5, 2, 4 \to 5, 4 \to 5$ --- $4$ descending chains $5, 1, 2, 3, 4 \to 5, 2, 3, 4 \to 5, 3, 4 \to 5, 4 \to 5$ --- $4$ descending chains There are a total of $73$ descending chains in $24$ permutations, thus, $\boxed{A(5) = \frac{73}{24}}$. Also, $S(5) = A(1) + A(2) + A(3) + A(4) + A(5) = 1 + \frac{3}{2} + \frac{7}{3} + \frac{73}{24} = \frac{63}{8}$. Plugging $k=5$ and $n \to n-1$ into $S(n) + \frac{n}{2} \ge \frac{n(n+1)}{k(k+1)}\left(S(k) + \frac{k}{2} \right)$, we get $$S(n-1) \ge \frac{n(n-1)}{30}\left(\frac{63}{8} + \frac{5}{2}\right) - \frac{n-1}{2} = (n-1)\left(\frac{83n}{240} - \frac{1}{2}\right).$$Plugging this into $A(n) \ge \frac{2S(n-1)}{n-1} + \frac{1}{2}$, we get $$A(n) \ge \frac{83n}{120}-1+\frac{1}{2} = \frac{83n}{120}-\frac{1}{2}.$$ Plugging $k=5$ and $n \to n-1$ into $S(n) + n - \frac{1}{4} \le \frac{n(n+1)}{k(k+1)}\left(S(k) + k - \frac{1}{4} \right)$, we get $$S(n-1) \le \frac{n(n-1)}{30}\left(\frac{63}{8}+5-\frac{1}{4}\right)-(n-1)+\frac{1}{4}=(n-1)\left(\frac{101n}{240}-1\right)+\frac{1}{4}.$$Plugging this into $A(n) \le \frac{2S(n-1)}{n-1} + \frac{2n-3}{2n-2}$, we get $$A(n) \le \frac{101n}{120} - 2 + \frac{1}{2n-2} + \frac{2n-3}{2n-2} = \frac{101n}{120} - 1 < \frac{101n}{120} - \frac{1}{2}.$$Therefore, $\frac{83n}{120}-\frac{1}{2} \le A(n) \le \frac{101n}{120} - \frac{1}{2}$.