Let $\{a_n\}$ and $\{b_n\}$ and $\{c_n\}$ be sequences such that $a_n \leq b_n \leq c_n$ for all $n \geq N_0$, $N_0 \in \mathbb N$. Suppose that $\{a_n\}$ and $\{c_n\}$ both converge to $L \in \mathbb R$. Prove that $\{b_n\}$ also converges to L.

My solution:

WTS: $\exists L \in R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in N$, if $n > N$, then $|b_n - L| < \text{ I don't know yet}$

Let $\epsilon > 0$ be arbitrary

Choose $N = \text{ I don't know yet}$

We know that $a_n \leq b_n \leq c_n$.

Suppose $n > N$, then...

I don't know how to do this. Could anyone point me to the right direction? Thx



$$\,\,\,\,\,\,|a_n - L | < \epsilon \implies -\epsilon < a_n - L \leqslant b_n - L \\ |c_n - L| < \epsilon \implies b_n - L \leqslant c_n -L < \epsilon$$

  • $\begingroup$ Writing all that down its kinda clear that $-\epsilon< b_n - L < \epsilon$, which is $|b_n - L| < \epsilon$ as required. Do I gotta go further and explain to complete the proof? Oh and thx this helped alot $\endgroup$ – user349557 Mar 22 '17 at 2:47
  • $\begingroup$ Finish by showing for any $\epsilon > 0$, there exist positive integers $N_1, N_2$ such that $n >N_1 \implies |a_n-L| < \epsilon$ and $n >N_2 \implies |c_n-L| < \epsilon.$ You already said as much. Now what if $n > \max(N_1,N_2)$? $\endgroup$ – RRL Mar 22 '17 at 2:51

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