If $a_n \to L$ and $b_n$ lies between $a_n$ and $a_{n+1}$ for each $n$, prove $b_n \to L$.

I know that we are not given which direction the inequalities go. So, there would be two cases:

  1. $a_n \leq b_n \leq a_{n+1}$, and

  2. $a_n \geq b_n \geq a_{n+1}$.

Intuitively, I would want to use the Squeeze Theorem to show that there is a sequence that is bounded below for the first case and another one bounded above for the second case. However, how does one create such sequences without losing validity and still having $b_n$ to lie in between them? One argument I can think of is letting $M = \max{\{a_n,}{a{n+1}\}}$ and similarly letting $m=\min{\{a_n,}{a_{n+1}\}}$. Is this acceptable or would it be mathematically incorrect?

Are there other ways to proceed in solving this problem?


We do a basic $\epsilon$-$N$ argument.

Let $\epsilon\gt 0$ be given. We show that there is an $N$ such that if $n\gt N$ then $|b_n-L|\lt \epsilon$.

Since the sequence $(a_n)$ has limit $L$, there is an $n$ such that if $n\gt N$ then $|a_n-L|\lt \epsilon/2$.

Also, if $n\gt N$, then $|a_{n+1}-L|\lt \epsilon/2$.

But by the Triangle Inequality, it follows that $|a_{n+1}-a_n|\lt \epsilon/2+\epsilon/2=\epsilon$. Since $b_n$ is between $a_n$ and $a_{n+1}$, we have $\min(|b_n-a_n|,|b_n-a_{n+1}|)\lt \epsilon/2$.

It follows from the Triangle Inequality that $|b_n-L|\lt\epsilon/2+\epsilon/2=\epsilon$.


  • $\begingroup$ Ahh the $\epsilon-N$ argument. I know it as the $K-\epsilon$ principle as per my textbook, but it is the same concept. Very concise response, many thanks for this @Andre Nicolas. $\endgroup$ – Jamil_V Sep 17 '13 at 2:07
  • $\begingroup$ You are welcome. This argument uses no "facts" about limits. If you have already proved a few basic theorems, there are easier arguments. $\endgroup$ – André Nicolas Sep 17 '13 at 2:11

I would argue that because $lim_{n\to\infty} a_n = L_1$ and $lim_{n\to\infty} a_{n+1} =L_2$, and $L_1=L_2$, you know that, for any arbitrary sequence $b_n$ fitting the qualification, $lim_{n\to\infty} b_n =L$ with $L_1\le L\le L_2$. I didn't really add much formality, but I hope that'll help.


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