The first terms of the sequence are: $a_1=0, a_2=\frac{1}{2}, a_3=\frac{7}{12}, a_4=\frac{1}{2}, a_5=\frac{101}{180}, a_6=\frac{19}{36}, a_7=\frac{181}{336}$ and so on. By simple algebra we can verify that $a_3$ is the biggest among all, so now we will prove for all $n>3$ that $a_n<\frac{7}{12}$ by strong induction, with the previous info we have and base cases $n=4,5,6,7$, so now we focus on $n \ge 8$, suppose that it holds for all $n \le \ell-1$, we will prove it for $\ell$, we now split in cases: Case 1: $\ell=2k$ $$a_{2k}<\frac{1}{2k}+\frac{1}{k} \cdot \left(\frac{19+(k-4)7}{12} \right)=\frac{1}{2k}+\frac{1}{k} \cdot \frac{7k-9}{12}=\frac{1}{2k}+\frac{7}{12}-\frac{9}{12k}<\frac{7}{12}$$Case 2: $\ell=2k+1$ $$a_{2k+1}<\frac{1}{2k+1}+\frac{1}{k+1} \cdot \left( \frac{19+((k+1)-4)7}{12} \right)=\frac{1}{2k+1}+\frac{1}{k+1} \cdot \frac{7(k+1)-9}{12}=\frac{1}{2k+1}+\frac{7}{12}-\frac{9}{12(k+1)}<\frac{7}{12}$$Therefore the induction is complete so we indeed have that the maximal term in the sequence is $a_3=\frac{7}{12}$ thus we are done .