The answer is $\boxed{M_1=1/2}$, $\boxed{M_2=7/6}$ and $\boxed{M_n=n-1}$ for all $n\ge 3$. First case: $n=1$ \[a_1\ge 2\quad\implies\quad \frac{1}{a_1}\le \frac{1}{2}\] Second case: $n=2$ Obviously the expression is maximal when $a_2=a_1+1$ so \[\frac{1}{a_1}+\frac{a_1}{a_1+1}=\frac{a_1^2+a_1+1}{a_1^2+a_1}=1+\frac{1}{a_1^2+a_1}\le 1+\frac{1}{4+2}=\frac{7}{6}\]Equality holds for $a_1=2$ and $a_2=3$. General case: $n-1$ can be achieved For $n\ge 3$ first we prove that the expression can get arbitrarily close to $n-1$. Take $a_i=N+i$ and let $N\rightarrow\infty$. The first fraction goes to $0$ while all the others go to $1$, so the expression goes to $n-1$. General case: $n-1$ is an upper bound We will prove that \[\frac{1}{a_1}+\frac{a_1}{a_2}+\frac{a_2}{a_3}\le\frac{1}{a_1}+\frac{a_1}{a_2}+\frac{a_2}{a_2+1}\le 2\]Since all the other $n-3$ terms are $<1$, this proves that the whole expression is $<n-1$. The inequality rewrites as \[\frac{a_2^2+a_2+a_1^2a_2+a_1^2+a_2^2a_1}{a_1a_2^2+a_1a_2}\le 2\iff a_1a_2(a_2-a_1)\ge(a_2-a_1)^2+a_2\]Which is true because \[a_1a_2(a_2-a_1)\ge a_2(a_2-a_1)+a_2(a_2-a_1)\ge (a_2-a_1)^2+a_2\]