Thread: Limit of two sequences

Let a_n and b_n be two convergent sequences. Prove: if for every even n: a_n<=b_n, and for every odd n: a_n>=b_n, then lim a_n = lim b_n.

Any ideas? I'm looking for a formal proof.
Thanks!

This can be proved using the following facts.

(1) The limit of a subsequence of a converging sequence equals the limit of the sequence.

(2) If a_n and b_n are converging sequences and a_n ≤ b_n for all n, then lim a_n ≤ lim b_n.

The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
"If a sequence converges, then its limit is unique."
"Every convergent sequence is bounded."

Let A = lim aₙ and B = lim bₙ. For a given ε > 0, choose N such that for all n > N we have |aₙ - A| < ε and |bₙ - B| < ε. Pick any even n > N; then inequalities in the previous sentence imply that A < aₙ + ε and bₙ < B + ε. Therefore, A < aₙ + ε ≤ bₙ + ε < B + 2ε. Thus, for the given ε we showed that A < B + 2ε. Since this holds for any positive ε, it must be that A ≤ B. Similarly, you can show that A ≥ B.

Originally Posted by loui1410

The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
"If a sequence converges, then its limit is unique."
"Every convergent sequence is bounded."

Suppose that . Going for a contradiction, suppose that . Now define .
So we have .

Because of convergence, almost all the are in and almost all the are in .
But that is impossible because of the alteration of odd and even terms.

emakarov, thank you - that's a great solution, I wouldn't have thought about it.
Plato's solution, however, seems more natural to me as it's similar to some other examples I've already seen.
Thank you both.