For each $i=1,2,\dotsc ,n$ let $s_i$ denote partial sum $a_1+a_2+...+a_i$. For convenient, let $S=\sum_{i=1}^{n-1}{\frac{a_i}{a_{i+1}}}$. The given condition gives us $a_{i+1}\geqslant s_i$ for all $i=1,2,\dotsc ,n-1$. We then have \begin{align*} S& =\frac{s_1}{a_2}+\sum_{i=2}^{n-1}{\frac{s_i-s_{i-1}}{a_{i+1}}}\\ & =\left( \sum_{i=1}^{n-2}{s_i\left(\frac{1}{a_{i+1}} -\frac{1}{a_{i+2}}\right)} \right)+\frac{s_{n-1}}{a_n}\\ & \leqslant \left( \sum_{i=1}^{n-2}{a_{i+1}\left(\frac{1}{a_{i+1}} -\frac{1}{a_{i+2}}\right)} \right) +\frac{a_n}{a_n}\\ & = (n-1)-\sum_{i=1}^{n-2}{\frac{a_{i+1}}{a_{i+2}}}\\ & = (n-1)+\frac{a_1}{a_2}-S \leqslant n-S. \end{align*}Note that the last inequality follows from $a_2\geqslant a_1$. In conclusion, we get $2S\leqslant n\implies S\leqslant \frac{n}{2}$. Equality holds, for example, at $a_1=1$ and $a_i=2^{i-2}$ for all $i=2,3,\dotsc ,n$.

For each $i=1,2,...,n-1$ let $x_i=\frac{s_i}{a_{i+1}}$, where $s_i=a_1+a_2+...+a_i$ . The inequality becomes: $$S=x_1+\frac{x_2}{1+x_1}+\cdots \frac{x_{n-1}}{1+x_{n-2}} \le \frac{n}{2}$$Now if we look at $S$ as a function of $x_i$, by convexity we see that the maximum value of $S$ is achieved when all $x_i$s are equal to $1$.