Bounded Sequences
Sequences can be bounded from above and from below. The values of a bounded sequence cannot become arbitrarily large.
Discover Bridges
Study Bounded Sequences
Definition 1 (Boundedness of sequences). Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in \(\mathbb{R}\). Then \((a_n)_{n\in\mathbb{N}}\) is called
bounded from above if there exists some \(c\in\mathbb{R}\) with \(a_n \leq c\) for all \(n\in\mathbb{N}\).
bounded from below if there exists some \(c\in\mathbb{R}\) with \(a_n \geq c\) for all \(n\in\mathbb{N}\).
bounded if there exists some \(c\in\mathbb{R}\) such that for all \(n\in\mathbb{N}\) holds \(|a_n|\leq c\).
unbounded if it is not bounded, i.e., for all \(c\in\mathbb{R}\), there exists some \(n\in\mathbb{N}\) with \(|a_n|> c\).
Remark 2. It can be readily seen from the definition that a sequence is bounded if and only if it is both bounded from above and bounded from below.
Theorem 3. Let \((a_n)_{n\in\mathbb{N}}\) be a convergent sequence in \(\mathbb{R}\). Then \((a_n)_{n\in\mathbb{N}}\) is bounded.
Proof. Suppose that \(\lim_{n\to\infty}a_n=a\).
Take \(\varepsilon=1\). Then there
exists some \(N\) such that for all
\(n\geq N\) holds \(|a_n-a|<1\).
Thus, for all \(n\geq N\) holds \[|a_n|=|a_n-a+a|\leq
|a_n-a|+|a|<1+|a|.\]
Now choose \[c=\max\{|a_1|,|a_2|,\ldots,|a_{N-1}|,|a|+1\}\]
and consider some arbitrary sequence element \(a_k\).
If \(k<N\), then \(|a_k|\leq \max\{|a_1|,|a_2|,\ldots,|a_{N-1}|\}\leq
c\).
In the case \(k\geq N\), the above
calculations lead to \(|a_k|<|a|+1\leq
c\).
Altogether, this implies that \(|a_k|\leq
c\) for all \(k\in\mathbb{N}\),
so \((a_n)_{n\in\mathbb{N}}\) is
bounded by \(c\). ◻
Remark 4. For a convergent sequence \((a_n)_{n\in\mathbb{N}}\) that is bounded by \(c\), we can also deduce from the above argumentation that for the limit \(a\) holds \(|a|\leq c\):
Suppose that \(\lim_{n\rightarrow\infty} a_n = a\in\mathbb{R}\), then for an arbitrary \(\varepsilon>0\) there exists an \(N\in\mathbb{N}\) such that \(|a-a_n|<\varepsilon\) for all \(n\geq N\). Hence \(|a|=|a-a_n+a_n|\leq|a-a_n|+|a_n|\leq \varepsilon + c\). Since \(\varepsilon>0\) can be chosen arbitrarily small this implies \(|a|\leq c\).
Discuss your questions by typing below.
Solve the WeBWorK Exercise
The data for the interactive network on this webpage was generated with pntfx Copyright Fabian Gabel and Julian Großmann. pntfx is licensed under the MIT license. Visualization of the network uses the open-source graph theory library Cytoscape.js licensed under the MIT license.