limit of quotient of two series Let $Y_{n} > 0$ for all $ n\in \mathbb{N} $, with $\sum{Y_{n}}= +\infty$.
If $\displaystyle\lim\limits_{n\rightarrow \infty}\frac{X_{n}}{Y_{n}}= a$ then $\displaystyle\lim\limits_{n\rightarrow\infty}\frac{X_{1}+X_{2}+X_{3}+\dots+X_{n}}{Y_{1}+Y_{2}+Y_{3}+\dots+Y_{n}}= a$
I need a idea for solution.
can i supose that $\lim\limits_{n\rightarrow\infty}Y_{n}$ exist ?
 A: $\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}$This observation is perhaps trivial, but I think it's useful to mention that this is just a different form of Stolz-Cesaro theorem.
Let us have a look at these two claims.

Claim 1: Let $(x_n)$ and $(y_n)$ be real sequences. Let $(y_n)$ be positive, strictly increasing and $\limti n y_n=+\infty$.
  If there exists a limit
  $$\limti n\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=L,$$
  then
  $$\limti n\frac{x_n}{y_n}=L.$$
Claim 2: Let $(a_n)$ and $(b_n)$ be real sequences such that $b_n>0$ for each $n\in\mathbb N$ and
  $\sum b_n=+\infty$. If there exists a limit
  $$\limti n\frac{a_n}{b_n}=L,$$
  then
  $$\limti n\frac{\sum_{k=1}^n a_k}{\sum_{k=1}^n b_k}=L.$$

By taking $a_n=x_{n+1}-x_n$ and $b_n=y_{n+1}-y_n$ we get Claim 2 from Claim 1. Conversely, by taking the partial sums $x_n=\sum_{k=1}^n a_n$ and $y_n=\sum_{k=1}^n b_n$ we get Claim 1 from Claim 2.

Basically with the same proof a slightly more general result can be shown.
The proof can be found e.g. here.

Theorem
  Let $(x_n)$ and $(y_n)$ be real sequences. Let $(y_n)$ be positive, strictly increasing and $\limti n y_n=+\infty$.
  $$
\liminf\frac{x_{n+1}-x_n}{y_{n+1}-y_n}\le
\liminf\frac{x_n}{y_n}\le \limsup\frac{x_n}{y_n}\le
\limsup\frac{x_{n+1}-x_n}{y_{n+1}-y_n}.
$$

A: it sounds like you'd like a less lazy hint, so here goes:
Fix $\epsilon > 0$, we'll show that the partial sums $\frac{X_1 + \ldots + X_n}{Y_1 + \ldots +Y_n}$ are eventually trapped in $[a-\epsilon, a+ \epsilon]$, which tells us exactly what we want.
Right, for that same $\epsilon$, we know that for $n > N$ we have $$a - \epsilon < \frac{X_n}{Y_n} < a + \epsilon$$so our partial sums are bounded by: $$\frac{(X_1 + \ldots + X_N) + (a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n} <  \frac{X_1 + \ldots + X_n}{Y_1 + \ldots +Y_n} < $$$$\frac{(X_1 + \ldots + X_N) + (a+\epsilon)Y_{N+1} + \ldots + (a + \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$(I just literally stuck in our estimate) Now, the idea is that the left and right hand bounds converge to $a - \epsilon$, $a + \epsilon$ respectively.
I'll just show one of them, $a - \epsilon$. First ``tear off" the first $N$ terms, since $\sum Y_n = \infty$ they'll dissappear in the limit, that is we have: $$\lim_n \frac{(X_1 + \ldots + X_N) + (a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n}  = $$$$\lim_n \frac{X_1 + \ldots + X_N}{Y_1 + \ldots + Y_n} + \lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$$$0 + \lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n}$$Right, now to manage this guy, notice that if you forgot about $Y_1 + \ldots + Y_N$ in the denominator, you'd just have $a-\epsilon$, which is what we want, and sure enough: $$\lim_n \frac{(a-\epsilon)Y_{N+1} + \ldots + (a - \epsilon)Y_n}{Y_1 + \ldots + Y_n} = \lim_n \frac{(a - \epsilon)Y_{N+1} + \ldots + (a-\epsilon)Y_n}{Y_{N+1} + \ldots + Y_n} \frac{Y_{N+1} + \ldots + Y_n}{Y_1 + \ldots + Y_n}$$$$ = (a-\epsilon) \lim_n \frac{Y_{N+1} + \ldots + Y_n}{Y_1 + \ldots + Y_n} = (a - \epsilon) \cdot 1$$which is what we wanted. 
Comment or something if you have any questions etc. I hope this helps mate :)
