# Two sequences approach to zero

If $x_n$ and $y_n$ are two sequences in $\mathbb R$, such that $x_n\longrightarrow 0$ and $y_n\longrightarrow 0$ as $n\longrightarrow \infty$. Is there any $N$ such that say $x_n\leq (or\ \geq) \ y_n$ for all $n\geq N$? Is there any property for that?

• If the sequences are (for example) $\cos n\over n$ and $\sin n\over n$, how would one ever find such an $N$? – abiessu Feb 24 '16 at 16:17
• The answer is negative. Take, for example, $x_n=(-1)^n/n$ and $y_n=(-1)^{n+1}/n$. – Cm7F7Bb Feb 24 '16 at 16:18
• What I want is $x_n/y_n$ or $y_n/x_n$ is converges to zero? is that always possible ? – Ronald Feb 24 '16 at 16:28
• If you have that and they are both positive, then yes. If they don't keep the same sign, they no, again... (Take $x_n = 1/n$ and $y_n = (-1)^n/n^2$, for instance.) – Clement C. Feb 24 '16 at 16:31

• Take $x_n = \frac{(-1)^n}{n}$ and $y_n = 0$.
• Even with constant sign: Take $x_n = \frac{2+(-1)^n}{n}$ and $y_n = \frac{2}{n}$.
• Even with monotonicity: take $x_n = \frac{1}{n}$ and $$y_n= \begin{cases} \frac{1}{3k+1} & \text{ if } n=3k\\ \frac{1}{3k+1} & \text{ if } n=3k+1\\ \frac{1}{3k+1} & \text{ if } n=3k+2\end{cases}$$ (i.e., $(y_n)_n$ is non-increasing, constant on consecutive blocks of 3 indices; $(x_n)_n$ is decreasing.)
• The last example can easily be modified to have strict monotonicity of $(y_n)_n$ as well. – Clement C. Feb 24 '16 at 16:23