# If $a_n \sim b_n$ and $a_n \leq c_n$, when does $b_n \leq c_n$ also?

Let $(a_n)$ and $(b_n)$ be positive sequences with $a_n \sim b_n$. Suppose also that for some other positive sequence $(c_n)$, I have $a_n \leq c_n$ for all $n$. Also suppose that $a_n$ and $b_n$ alternate only finitely many times.

I would like to say that: $$b_n \leq c_n \textrm{ eventually }$$ i.e. there exists $N$ such that $b_n \leq c_n$ for all $n \geq N$. Clearly the statement isn't true in general, for example taking $a_n = n$, $b_n = n+1$, and $c_n = n$. So what is the weakest assumption I can impose to guarantee this holds in general?

The specific case I'm looking at is where all the sequences $a_n, b_n, c_n$ are $\in (0,1)$ and increasing to $1$. But I'm curious about the general case too.

• It may be hard to find a useful formulation, even in your specific case. Take $c_n=a_n$, and let $b_n$ alternately be ahead of and behind $a_n$. – André Nicolas Oct 13 '15 at 15:08
• @AndréNicolas good point. maybe for simplicity I will suppose that $a_n$ and $b_n$ alternate only finitely many times. – gogurt Oct 13 '15 at 15:11
• $b_{n} = o(a_{n})$ as $n \to \infty$. – Megadeth Oct 13 '15 at 15:14
• @GudsonChou but $a_n/b_n \to 1$, so... – gogurt Oct 13 '15 at 15:18
• Two simple extra conditions are $$\liminf_{n \to \infty} \frac{c_n}{a_n} > 1$$ or $$\limsup_{n \to \infty} \frac{b_n - a_n}{c_n - a_n} < \infty,$$ but maybe they aren't very useful. You probably won't be able to get a condition which just uses asymptotic notation since it isn't fine-grained enough for this kind of thing. – Antonio Vargas Oct 13 '15 at 15:52