65. Let $a_{n+1}=a_{1}, S_{n}=\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{n}+a_{1}}$, then
$$S_{n}>\frac{a_{1}}{a_{1}+a_{2}+\cdots+a_{n}}+\frac{a_{2}}{a_{1}+a_{2}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n}}=1$$

Also,
$$$$$\text{S n +1= S n + a 1 a 1 +a 2 + +a n + a 2 a 1 +a 2 + +a n + + a n a 1 +a 2 + +a n 0}$$, then we have$$S_{n}=\frac{n-1}{1+t}+\frac{t^{n-1}}{1+t^{n-1}}\$.$
Since $\lim _{t \rightarrow 0} S_{n}(t)=n-1, \lim _{t \rightarrow+\infty} S_{n}(t)=1$, the bounds of the inequality are optimal.