Theorem 3.3.1.
Let \(a\lt b\text{.}\) Suppose \(c_n\in [a,b]\) for all \(n\text{,}\) and \(\displaystyle \lim_{n\rightarrow \infty}c_n=c\text{.}\) Then \(c\in [a,b]\text{.}\)
Section 3.3 Some basic theorems
Subsection 3.3.1 Sequences in closed intervals
The next theorem is one of the reasons why an interval \([a,b]\) is called “closed” (if a sequence in in \([a,b]\text{,}\) its limit cannot “get out” of \([a,b]\)).
Hint.
Proceed by contradiction and assume for example that \(c\gt b\text{.}\) Then consider \(\epsilon=(c-b)/2\text{.}\)
Subsection 3.3.2 The squeeze theorem
This theorem is so popular that has its own name.
Theorem 3.3.2. The squeeze theorem.
Suppose \((a_n)\text{,}\) \((b_n)\) are sequences, both converging to the same limit \(L\text{.}\) If \((c_n)\) is a sequence such that \(a_n\leq c_n\leq b_n\) for all \(n\text{,}\) then \((c_n)\) converges to \(L\text{.}\)
The name of the theorem should be clear: if the two sequences \((a_n),(b_n)\) both approach the same value \(L\) and \((c_n)\) is “squeezed” between them, then \((c_n)\) will also approach \(L\text{.}\) See the figure below,
Hint.
If \(n_0\) “works” for both \((a_n)\) and \((b_n)\text{,}\) show that it also “works” for \((c_n)\text{.}\)
The next example is probably the simplest application of the squeeze theorem, and it is often used in many problems involving the sine and cosine functions. While we have not yet defined what \(\sin\) and \(\cos\) are in this course, we will rely on previous knowledge of these functions from the Precalculus or Calculus courses. In particular, we will use the fact that the range of \(\sin x\) and \(\cos x\) is \([-1,1]\text{.}\)
Example 3.3.4.
Let \(\displaystyle a_n=\frac{\sin n}{n}\text{.}\) Since \(-1\leq \sin n\leq 1\text{,}\) we have \(|\sin n|\leq 1\text{,}\) and so
\begin{equation*}
0\leq |a_n|= \frac{|\sin n|}{n}\leq \frac{1}{n}.
\end{equation*}
Since the constant sequence \(0\) and the sequence \(1/n\) both converge to \(0\text{,}\) we conclude from the squeeze theorem that \(\displaystyle \lim_{n\rightarrow \infty} a_n=0\text{.}\)
Subsection 3.3.3 A bounded, monotone sequence converges
We are now ready to discuss a result that is crucial to Real Analysis. We will often use it in this course to prove that a sequence converges.
Theorem 3.3.5.
Suppose that \((a_n)\) is monotone and bounded. Then \((a_n)\) converges.
Hint.
Proof Assume \((a_n)\) increasing. Given \(\epsilon \gt 0\text{,}\) \(a-\epsilon\) cannot be an upper bound for the sequence, so there must be some \(n_0\in {\mathbb N}\) such that \(a_{n_0}\gt a-\epsilon\text{.}\) If \(n\geq n_0\text{,}\) since \((a_n)\) is increasing, and \(s\) is an upper bound, we find \(a-\epsilon\lt a_{n_0} \leq a_n \leq a\lt a+\epsilon\text{,}\) that is, \(|a_n-a|\lt \epsilon .\)
