Property of $\frac{\sum a_i}{\sum b_i}$ when $\frac{a_i}{b_i}$ is increasing Can you prove the following property?
Let $a_i,b_i$ be real positive numbers for $i=1,2,\dots$ , such that $\dfrac{a_i}{b_i}$ is an increasing sequence. Then:
1) The sequence $s_n=\dfrac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}$ is increasing.
2) Furthermore, for $0<r<k<n$ we have:
$$\dfrac{\sum_{i=1}^r a_i}{\sum_{i=1}^r b_i}<\dfrac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}<\dfrac{\sum_{i=k}^n a_i}{\sum_{i=k}^n b_i}$$
 A: If $b_1b_2\gt0$ and $\dfrac{a_1}{b_1}\leqq\dfrac{a_2}{b_2}$, then adding $a_1b_1$ to both sides yields
$$
\begin{align}
a_1b_2&\leqq a_2b_1\\
a_1(b_1+b_2)&\leqq b_1(a_1+a_2)\\
\frac{a_1}{b_1}&\leqq\frac{a_1+a_2}{b_1+b_2}\tag{1}
\end{align}
$$
and adding $a_2b_2$ to both sides yields
$$
\begin{align}
a_1b_2&\leqq a_2b_1\\
(a_1+a_2)b_2&\leqq a_2(b_1+b_2)\\
\frac{a_1+a_2}{b_1+b_2}&\leqq\frac{a_2}{b_2}\tag{2}
\end{align}
$$
Therefore,
$$
\frac{a_1}{b_1}\leqq\frac{a_1+a_2}{b_1+b_2}\leqq\frac{a_2}{b_2}\tag{3}
$$
where $\leqq$ means that if $\lt$ holds in the hypothesis, $\lt$ holds in the conclusion, and if $=$ holds in the hypothesis, $=$ holds in the conclusion.
Using $(3)$ and induction gives the desired results.
Suppose that
$$
\frac{\displaystyle\sum_{k=1}^{n-1}a_k}{\displaystyle\sum_{k=1}^{n-1}b_k}
\leqq\frac{a_{n-1}}{b_{n-1}}
\leqq\frac{a_n}{b_n}\tag{4}
$$
then $(3)$ and $(4)$ give
$$
\frac{\displaystyle\sum_{k=1}^{n-1}a_k}{\displaystyle\sum_{k=1}^{n-1}b_k}
\leqq\frac{\displaystyle\sum_{k=1}^{n}a_k}{\displaystyle\sum_{k=1}^{n}b_k}
\leqq\frac{a_n}{b_n}\tag{5}
$$
The other inequalities are proven similarly.
