Show that $\lim_{n \to +\infty} \frac{\sum^n x_i}{\sum^n y_i}=a.$ Let $\{y_n\}$ a sequence so that $\sum y_i=+\infty$ and
$$y_n>0, \forall n \in \mathbb{N}.$$
Show that if 
$$\lim_{n \to +\infty} \frac{x_n}{y_n}=a$$
then
$$\lim_{n \to +\infty} \frac{\displaystyle \sum^n x_i}{\displaystyle \sum^n y_i}=a.$$
 A: Let $\epsilon >0$. Then there exists some $N$ so that for all $n > N$ we have
$$\left| \frac{x_n}{y_n} -a \right|<\epsilon$$
Therefore 
$$\left| x_n -ay_n\right| < \epsilon y_n \,.$$
As $y_n$ is positive, it follows that 
$$\sum_{i=N}^n \left| x_i -ay_i\right| \leq \sum_{i=N}^n \epsilon y_i \leq \sum_{i=1}^n \epsilon y_i= | \sum_{i=1}^n \epsilon y_i| (*)$$
Therefore, for all $n >N$ we have
$$\left|\frac{\displaystyle \sum_{i=1}^n x_i}{\displaystyle\sum_{i=1}^n  y_i}-a \right| \leq\left|\frac{\displaystyle \sum_{i=1}^N x_i}{\displaystyle\sum_{i=1}^n  y_i} \right|+ \left|\frac{\displaystyle \sum_{i=N}^n x_i}{\displaystyle\sum_{i=1}^n  y_i}-a \right|\\
=\left|\frac{\displaystyle \sum_{i=1}^N x_i}{\displaystyle \sum_{i=1}^n  y_i} \right|+ \left|\frac{\displaystyle \sum_{i=N}^n x_i-\displaystyle \sum_{i=1}^n a
y_i}{\displaystyle \sum_{i=1}^n  y_i} \right| \\
\leq \left|\frac{\displaystyle \sum_{i=1}^N x_i}{\displaystyle \sum_{i=1}^n  y_i} \right|+ \left|\frac{\displaystyle \sum_{i=N}^n (x_i- a
y_i)}{\displaystyle\sum_{i=1}^n  y_i} \right|+\left|\frac{\displaystyle \sum_{i=1}^N  a
y_i}{\displaystyle\sum_{i=1}^n  y_i} \right|$$
Now  $(*)$ tells us that the middle term is $<\epsilon$. 
Since $N$ is fixed, and $\sum y_n \to \infty$ it is easy to prove that
$$\lim_n\frac{\displaystyle \sum_{i=1}^N x_i}{\displaystyle\sum_{i=1}^n  y_i} = \lim_n \frac{\displaystyle \sum_{i=1}^N  a
y_i}{\displaystyle\sum_{i=1}^n  y_i}  =0$$
Therefore, you can find some $N_1$ so that for all $n >N_1$ the first and third term in the above inequality are less that $\epsilon$. This completes the proof.
