Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be sequences such that $a_n ≤ b_n ≤ c_n$ for all $n ≥ N_0$, $N_0 ∈ \mathbb{N}$. Suppose that $\{a_n\}$ and $\{c_n\}$ both converge to $l ∈ \mathbb{R}$. Prove that $\{b_n\}$ also converges to $l$.

I am in Grade 12 and doing some self-learning with sequences. Could someone please let me know how to prove this?

I was thinking to construct $l - e < a_n \le b_n \le c_n < l + e$.

Do I have to "choose" $|a_n - l | <$ some epsilon and $|c_n - l| <$ some epsilon?


  • $\begingroup$ Exactly. And you have to tell for which n you have them smaller than $\varepsilon$. $\endgroup$ – Friedrich Philipp Mar 11 '16 at 4:59

Right. That's pretty much how you would go about. In detail, first fix $\varepsilon>0$. Then find $N_{1}$ and $N_{2}$ such that $|a_{n}-l|<\varepsilon$ and $|c_{k}-l|<\varepsilon$ for all $n\geq N_{1}$ and $k\geq N_{2}$. Choose $N_{\varepsilon}=\max\{N_{0},N_1,N_2\}$. Then \begin{align*} b_{n}-l\leq c_{n}-l\leq |c_{n}-l|<\varepsilon \end{align*} and \begin{align*} l-b_{n}\leq l-a_{n}\leq |l-a_{n}|<\varepsilon \end{align*} for all $n\geq N_{\varepsilon}$, then \begin{align*} |b_{n}-l|<\varepsilon \end{align*} for all $n\geq N_{\varepsilon}$. So $b_n \to l$.

  • $\begingroup$ @Eskin When do I choose the Max and when do I choose the min? $\endgroup$ – user1799252 Mar 11 '16 at 6:26
  • $\begingroup$ @user1799252. We only choose max here because we know that certain conditions of our interest are true for each of the cases $n\geq N_0$, $k\geq N_1$, and for $j\geq N_2$. Now we want all these conditions to hold simultaneously, so if $n\geq \max\{N_0, N_1, N_2\}$ then certainly $n\geq N_0$ and $n\geq N_1$ and $n\geq N_2$. $\endgroup$ – T. Eskin Mar 11 '16 at 19:10

Hint: I find it useful to think of $\epsilon - N$ proofs as challenge-response. If you claim $b_n \to l$ I am allowed to challenge you with an $\epsilon$ and you have to give me an $N$ such that $|b_n-l| \lt \epsilon$ for all $n \gt N$. Now you were told by somebody that $a_n \to l$ and $c_n \to l$, so you get to challenge them with the same type of question. It doesn't have to be the same $\epsilon$, often you want to divide it by some number, but the same one works here. So ask them for an $M$ so that $|a_n - l| \lt \epsilon$ whenever $n \gt M$ and an $M'$ so that $|c_n - l| \lt \epsilon$ whenever $n \gt M'$ Now you can come back to me with (what?)


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