Suppose that $p_1, \dots, p_n$ are nonnegative real numbers such that $p_1 + \cdots + p_n = 1$; denote the corresponding set of vectors by $\Delta_n$.
I am interested in the following function, $f \colon \Delta_n \to \mathbb{R}_+$, given by $$ f(p) = \sum_{k=1}^n \frac{p_k}{\sum_{j=k}^n p_j}. $$ We always have $f(p) \leq n$, using the lower bound $\sum_{j\geq k} p_j \geq p_k$. However I feel this must be a loose bound on the quantity $$ \sup_{p \in \Delta_n} f(p), $$ since it requires that $\sum_{j >k} p_j = 0$ for all $k$ to be met with equality. Hence, I am wondering what the largest $f(p)$ can be when evaluated over the simplex?
Take $0<r<1$ and put $p_k=(r^{k-1}-r^k)/(1-r^n)$ for $1\leqslant k\leqslant n$. Then $$\sum_{k=1}^n\frac{p_k}{\sum_{j=k}^n p_j}=\sum_{j=1}^n\frac{r^{k-1}-r^k}{r^{k-1}-r^n}\underset{\color{gray}{\big[j=n-k+1\big]}}{=}\sum_{j=1}^n\frac{1-r}{1-r^j}.$$ This tends to $n$ as $r\to0$, thus $\sup_{p\in\Delta_n}f(p)=n$.