Let a_{n}and b_{n}be two convergent sequences. Prove: if for every evenn: a_{n}<=b_{n, }and for every oddn: a_{n}>=b_{n,}then lim a_{n}= lim b_{n}.

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

Thanks!

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- Jul 9th 2012, 03:24 AMloui1410Limit 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! - Jul 9th 2012, 04:14 AMemakarovRe: 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 a_{n}and b_{n}are converging sequences and a_{n}≤ b_{n}for all n, then lim a_{n}≤ lim b_{n}. - Jul 10th 2012, 06:56 AMloui1410Re: 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, 07:17 AMemakarovRe: 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, 07:33 AMPlatoRe: Limit of two sequences
Suppose that $\displaystyle {\lim _{n \to \infty }}{a_n} = A\;\& \;{\lim _{n \to \infty }}{b_n} = B$. Going for a contradiction, suppose that $\displaystyle A<B$. Now define $\displaystyle \varepsilon = \frac{{B - A}}{2} > 0$.

So we have $\displaystyle \left( {A - \varepsilon ,A + \varepsilon } \right) \cap \left( {B - \varepsilon ,B + \varepsilon } \right) = \emptyset$.

Because of convergence,the $\displaystyle a_n's$ are in $\displaystyle \left( {A - \varepsilon ,A + \varepsilon } \right)$ and*almost all*the $\displaystyle b_n's$ are in $\displaystyle \left( {B - \varepsilon ,B + \varepsilon } \right)$.*almost all*

But that is impossible because of the alteration of odd and even terms. - Jul 10th 2012, 07:49 AMloui1410Re: 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.