$\lim \frac{x_{n+1}-x_n}{y_{n+1}-y_n}=a$ $\implies$ $\lim\frac{x_n}{y_n}=a$

Question

I would like to know how to solve this question:

If $(y_n)$ is increasing and $\lim y_n=+\infty$, then $\lim \frac{x_{n+1}-x_n}{y_{n+1}-y_n}=a$ $\implies$ $\lim\frac{x_n}{y_n}=a$

I don't know even how to begin, I really need help

Thanks a lot

Answer 1

We present the proof:

Rewrite $\dfrac{x_{n+1} - x_n}{y_{n+1}-y_n} \to a$ as saying that for sufficiently large $n$,

$$a-\epsilon < \dfrac{x_{n+1} - x_n}{y_{n+1}-y_n} < a + \epsilon$$

Since $y_n$ is increasing, we can multiply on both sides by the denominator of the middle to get

$$(a- \epsilon)(y_{n+1} - y_n) < x_{n+1} - x_n < (a+\epsilon)(y_{n+1} - y_n)$$

Sum this relation up to $K$, where we will specify $K$ later, and we get

$$(a - \epsilon)\sum_{n = n_0}^{K} (y_{n+1} - y_n) < \sum_{n = n_0}^K (x_{n+1} - x_n) < (a + \epsilon)\sum_{n = n_0}^{K} (y_{n+1} - y_n)$$

which is the same as

$$(a - \epsilon)(y_{K+1} - y_{n_0}) < x_{K+1} - x_{n_0} < (a + \epsilon)(y_{K+1} - y_{n_0})$$

This is a good place to be. As we know $y_n \to \infty$, let $K$ be at least large enough for $y_k$ to be positive for $k > K$. Then divide by $y_{K+1}$ everywhere and rearrange to get

$$(a - \epsilon)\left(1 - \frac{y_{n_0}}{y_{K+1}}\right) + \frac{x_{n_0}}{y_{K+1}} < \frac{x_{K+1}}{y_{K+1}} < (a + \epsilon)\left(1 - \frac{y_{n_0}}{y_{K+1}}\right) + \frac{x_{n_0}}{y_{K+1}}$$

$y_k \to \infty$, so for sufficiently large $K$ the bad fractions become arbitrarily small and we see that

$$\lim_{n \to \infty} \frac{x_n}{y_n} = a$$