Combinatorial problem on sum of integers

Question

Let $\mathcal{A}_n$ be the collection of all sequences of positive integers $A_n:=\langle n_1, n_2, \ldots, n_k\rangle$, where $n:=n_1$, $~2\le k\le n-1$ and $n_1>n_2>\ldots>n_k$. We denote by $S(A_n)$ the sum $\sum_{i=1}^{k-1} \frac{n_i}{n_{i+1}}$, and by $A^*_n$ the sequence of $\mathcal{A}_n$ maximizing this sum.

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Question: What is $S^*_n:=S(A^*_n)$?

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Note: I think that the maximum for $S(A_n)$ is attained by the sequence $B_n:=\langle n, n-1, n-2, \ldots, 1\rangle$. If this is true, how can we prove that $A^*_n=B_n$?

Answer 1

Suppose $A^*_n$ has successive terms $a,b$ with $a>b+1$. Consider the effect of inserting the integer $a-1$ into the sequence.

It is clear that $$\frac{a}{a-1}+\frac{a-1}{b}>\frac{a}{b}$$ and so $S$ has been increased. Similarly, we can increase $S$ by extending the sequence to $1$, if necessary.

Hence $A^*_n=B_n$, as you suspected.