1. ## a<c<b

Ok, so {a},{b}, and {c} are sequences such that {a} converges to A, {b} converges to A and a<c<b for all n. prove that {c} converges to A.

I know I need arbitrary epsilon and arbitrary N, but we have 2 different N's. The 2 different N's confuse me. How do I get started?

2. Find an N that works for both sequences.
$n \geqslant N \Rightarrow \quad \left| {a_n - A} \right| < \varepsilon \,\& \,\left| {b_n - A} \right| < \varepsilon$.
$a_n \leqslant c_n \leqslant b_n \Rightarrow \quad - \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon$
Carry on.

3. Originally Posted by Plato
Find an N that works for both sequences.
$n \geqslant N \Rightarrow \quad \left| {a_n - A} \right| < \varepsilon \,\& \,\left| {b_n - A} \right| < \varepsilon$.
$a_n \leqslant c_n \leqslant b_n \Rightarrow \quad - \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon$
Carry on.
But I thought you ahd to do it with N1 nad N2 since they're different sequences. hence different N's.

4. Originally Posted by kathrynmath
But I thought you ahd to do it with N1 nad N2 since they're different sequences. hence different N's.
Let $N=N_1 + N_2$.
Then N works for both sequences.

5. Originally Posted by Plato
Find an N that works for both sequences.
$n \geqslant N \Rightarrow \quad \left| {a_n - A} \right| < \varepsilon \,\& \,\left| {b_n - A} \right| < \varepsilon$.
$a_n \leqslant c_n \leqslant b_n \Rightarrow \quad - \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon$
Carry on.
Ok, so I assume a<c<b and show it works?

6. Originally Posted by kathrynmath
Ok, so I assume a<c<b and show it works?
From $- \varepsilon < a_n - A \leqslant c_n - A \leqslant b_n - A < \varepsilon$
it comes at once $\left| {c_n - A} \right| < \varepsilon$.