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.
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, almost all the $\displaystyle a_n's$ are in $\displaystyle \left( {A - \varepsilon ,A + \varepsilon } \right)$ and almost all the $\displaystyle b_n's$ are in $\displaystyle \left( {B - \varepsilon ,B + \varepsilon } \right)$.
But that is impossible because of the alteration of odd and even terms.