Let $\{u_n\}$, $\{v_n\}$ real sequences such that $0<u_n\le v_n$ for all $n$. Then $$\lim_{n\to\infty}\frac{u_1+u_2+...+u_n}{v_1+v_2+...+v_n}=\lim_{n\to\infty}\frac{\frac{u_1}{v_1}+\frac{u_2}{v_2}+...+\frac{u_n}{v_n}}{n}$$
I have some problem with the last sentence. I can prove that, if $\lim_{n\to\infty}\frac{u_n}{v_n}$ exists, then both of the above limits exist and they are equal. But what happen in the case $\lim\frac{u_n}{v_n}\notin\mathbb{R}$ (it is clear that this limit can not be $\infty$.)?
Consider:
$u_n = \begin{cases} 1 & n\;\text{odd}\\ 2 & n\;\text{even}\end{cases}$
$v_n = \begin{cases} 10 & n\;\text{odd}\\ 2 & n\;\text{even} \end{cases}$
Then, $\lim\limits_{n\to\infty} \frac{u_1 + u_2 + \dots + u_n}{v_1 + v_2 + \dots + v_n} = \frac{1}{4}$.
However, $\lim\limits_{n\to\infty} \frac{\frac{u_1}{v_1} + \frac{u_2}{v_2} + \dots + \frac{u_n}{v_n}}{n} = \frac{\frac{1}{10} + 1}{2} = \frac{11}{20} \neq \frac{1}{4}$.
So this statement appears to be false without further assumptions on $u_n$ and $v_n$.