does $\lim_{N\to\infty}\frac{\sum_{i=1}^N a_i}{\sum_{i=1}^N b_i}$ converge to $\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$ Can this ever be the case? 
$$\lim\limits_{N\to\infty}\frac{\sum\limits_{i=1}^N a_i}{\sum\limits_{i=1}^N b_i} = \lim\limits_{N\to\infty}\frac{1}{N}\sum_{i=1}^N\frac{a_i}{b_i}$$ with $a_i>0$, $b_i>0$, $a_i<b_i$. As others pointed out simulations indicate convergence, but is there formal ground to it?
 A: For a simple counterexample, take $a_i=\dfrac1{3^i}$ and $b_i=\dfrac1{2^i}$.
Then the LHS ratio tends to a finite value ($\frac12$), while the RHS tends to a finite value ($2$) over $N$, i.e. $0$.
A: Hint: try using Stolz theorem.
If the limit $\lim_{n\to\infty} \frac{a_n}{b_n}$ exists and is equal to $L$, then your limit is as well equal to $L$.
A: We want to show
\begin{equation}
  \lim\limits_{n\to\infty}\frac{\sum\limits_{i=1}^{n}{a_i}}{\sum\limits_{i=1}^{n}{b_i}}=\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i}
\end{equation}
Assuming right hand side is convergent, take
\begin{equation}
\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{i=1}^{n}\frac{a_i}{b_i}\to\bar{\varepsilon}
\end{equation}
rewrite as
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{a_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\bar\varepsilon
\end{equation}
let
\begin{equation}
\varepsilon_i = \frac{a_i}{b_i}
\end{equation}
multiply right hand side by 1 and substitute $\varepsilon_i$
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{\varepsilon_ib_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
substitute $\varepsilon_i=\bar\varepsilon+\Delta\varepsilon_i$ 
\begin{equation}
  \frac{\sum\limits_{i=1}^{\infty}{\bar\varepsilon b_i}+\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}=\frac{\sum\limits_{i=1}^{\infty}\bar\varepsilon{b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
Then it suffices to show 
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}\to 0
\end{equation}
Since $\forall i:a_i>0, b_i>0$ we obtain inequality
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<\frac{\left(\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}\right)\left(\sum\limits_{i=1}^{\infty}{b_i}\right)}{\sum\limits_{i=1}^{\infty}{b_i}}
\end{equation}
\begin{equation}
\frac{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i b_i}}{\sum\limits_{i=1}^{\infty}{b_i}}<{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}}
\end{equation}
where 
\begin{equation}
{\sum\limits_{i=1}^{\infty}{\Delta\varepsilon_i}}\to 0
\end{equation}
due to expectation $\mathrm{E}[\Delta\varepsilon_i]=0$ by its definition
