# Limit of two sequences

• Jul 9th 2012, 04:24 AM
loui1410
Limit of two sequences
Let an and bn be two convergent sequences. Prove: if for every even n: an<=bn, and for every odd n: an>=bn, then lim an = lim bn.

Any ideas? I'm looking for a formal proof.
Thanks!
• Jul 9th 2012, 05:14 AM
emakarov
Re: Limit of two sequences
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 an and bn are converging sequences and an ≤ bn for all n, then lim an ≤ lim bn.
• Jul 10th 2012, 07:56 AM
loui1410
Re: Limit of two sequences
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."
• Jul 10th 2012, 08:17 AM
emakarov
Re: Limit of two sequences
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.
• Jul 10th 2012, 08:33 AM
Plato
Re: Limit of two sequences
Quote:

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 ${\lim _{n \to \infty }}{a_n} = A\;\& \;{\lim _{n \to \infty }}{b_n} = B$. Going for a contradiction, suppose that $A. Now define $\varepsilon = \frac{{B - A}}{2} > 0$.
So we have $\left( {A - \varepsilon ,A + \varepsilon } \right) \cap \left( {B - \varepsilon ,B + \varepsilon } \right) = \emptyset$.

Because of convergence, almost all the $a_n's$ are in $\left( {A - \varepsilon ,A + \varepsilon } \right)$ and almost all the $b_n's$ are in $\left( {B - \varepsilon ,B + \varepsilon } \right)$.
But that is impossible because of the alteration of odd and even terms.
• Jul 10th 2012, 08:49 AM
loui1410
Re: Limit of two sequences
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.