The tasks sounds like this:
Given sequences $(x_n)_{n\in\mathbb{N}}$, $(y_n)_{n\in\mathbb{N}}$ in $\mathbb{R}$ where $\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=0$. It is also known, that $y_n$ is strong monoton. Show that when $$\frac{x_n}{y_n}$$ converges for $n\to\infty$, then it does not imply, that $$\frac{x_{n+1}-x_n}{y_{n+1}-y_n}$$ for $n\to\infty$ also converges
With a little help I got this:
- $a_n=\frac{x_n}{y_n}$, obviously $a_n\overset{n\to\infty}{\to}\alpha$
- $\frac{x_{n+1} – x_n}{y_{n+1} – y_n} = \frac{a_{n+1} y_{n+1} – a_n y_n}{y_{n+1} – y_n} = \frac{(a_{n+1} y_{n+1} – a_n y_{n+1}) + (a_n y_{n+1} – a_n y_n)}{y_{n+1} – y_n} = \frac{(a_{n+1} y_{n+1} – a_n y_{n+1})}{y_{n+1} – y_n} + a_n = \frac{y_{n+1} (a_{n+1} – a_n)}{y_{n+1} – y_n} + a_n$
- If we want to make this expression convergent, then we suppose, that:$$\frac{y_{n+1}(a_{n+1}-a_n)}{y_{n+1}-y_n}\to\beta$$
- Let $y=\frac{1}{n}$, then $$\frac{\frac{1}{n+1}(\frac{x_{n+1}}{\frac{1}{n+1}}-\frac{x_n}{\frac{1}{n}})}{\frac{1}{n+1}-\frac{1}{n}}=-n(n\cdot x_{n+1}+ x_{n+1}-n\cdot x_n)$$ and if we make $n$ infinitely big, we have $-\infty(\infty\cdot0+0-\infty\cdot0)$ what looks like bad to me.
What am I doing wrong?
Best Answer
Let $y_n={1\over n},$ $x_{2n}={1\over 2n}$ and $x_{2n+1}={1\over 2n}-{1\over 2(n+1)^4}.$ Then $y_n$ is strictly decreasing. By the way the $x_n$ is decreasing as well. The sequence ${x_n\over y_n}$ tends to $1.$ Moreover $$ {x_{2n+1}-x_{2n}\over y_{2n+1}-y_{2n}}$$ is convergent to $0, $ while $${x_{2n}-x_{2n-1}\over y_{2n}-y_{2n-1}}$$ tends to $1.$