Suppose that p 1 </msub> , &#x2026;<!-- … --> , p n </ms

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 ${\mathrm{\Delta }}_{n}$.

I am interested in the following function, $f:{\mathrm{\Delta }}_{n}\to {\mathbb{R}}_{+}$, given by
$f\left(p\right)=\sum _{k=1}^{n}\frac{{p}_{k}}{\sum _{j=k}^{n}{p}_{j}}.$
We always have $f\left(p\right)\le n$, using the lower bound $\sum _{j\ge k}{p}_{j}\ge {p}_{k}$. However I feel this must be a loose bound on the quantity
$\underset{p\in {\mathrm{\Delta }}_{n}}{sup}f\left(p\right),$
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 $f\left(p\right)$ can be when evaluated over the simplex?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Take $0 and put ${p}_{k}=\left({r}^{k-1}-{r}^{k}\right)/\left(1-{r}^{n}\right)$ for $1⩽k⩽n$. Then
$\sum _{k=1}^{n}\frac{{p}_{k}}{\sum _{j=k}^{n}{p}_{j}}=\sum _{j=1}^{n}\frac{{r}^{k-1}-{r}^{k}}{{r}^{k-1}-{r}^{n}}\underset{\left[j=n-k+1\right]}{=}\sum _{j=1}^{n}\frac{1-r}{1-{r}^{j}}.$
This tends to $n$ as $r\to 0$, thus $\underset{p\in {\mathrm{\Delta }}_{n}}{sup}f\left(p\right)=n$.