Bounded Monotonic Sequences
If a sequence of real numbers is bounded and monotonic, then it is convergent.
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Study Bounded Monotonic Sequences
Definition 1 (Monotonicity). A real sequence \((a_n)_{n\in\mathbb{N}}\) is called
monotonically increasing if for all \(n\in\mathbb{N}\) holds \(a_n\leq a_{n+1}\).
strictly monotonically increasing if for all \(n\in\mathbb{N}\) holds \(a_n < a_{n+1}\).
monotonically decreasing if for all \(n\in\mathbb{N}\) holds \(a_n\geq a_{n+1}\).
strictly monotonically decreasing if for all \(n\in\mathbb{N}\) holds \(a_n > a_{n+1}\).
We make essential use of Dedekind’s theorem to prove the following result:
Theorem 2 (Convergence of bounded and monotonic sequences). Let \((a_n)_{n\in\mathbb{N}}\) be a real sequence that has one of the following properties:
\((a_n)_{n\in\mathbb{N}}\) is monotonically increasing and bounded from above;
\((a_n)_{n\in\mathbb{N}}\) is monotonically decreasing and bounded from below;
Then \((a_n)_{n\in\mathbb{N}}\) is convergent.
Proof: Let us first assume that \((a_n)_{n\in\mathbb{N}}\) is monotonically
increasing and bounded from above. Define the set \(M=\{a_n\,:\,n\in\mathbb{N}\}\). Since \(M\) is bounded, Dedekind’s theorem implies
that there exists some \(K\in\mathbb{R}\) such that \[K=\sup M.\] We show that \(K\) is indeed the limit of the sequence
\((a_n)_{n\in\mathbb{N}}\).
Let \(\varepsilon>0\). By the
definition of the supremum, we have that \(a_n\leq K\) for all \(n\in\mathbb{N}\) and there exists some
\(N\in\mathbb{N}\) such that \(a_N>K-\varepsilon\). The monotonicity of
\((a_n)_{n\in\mathbb{N}}\) implies that
for all \(n\geq N\) holds \(a_N\leq a_n\). Altogether, we have \[K-\varepsilon<a_N\leq a_n\leq K\] and
thus \(|K-a_n|=K-a_n<\varepsilon\)
for all \(n\geq N\). This implies
convergence to \(K\).
To prove that convergence is also guaranteed in the case where \((a_n)_{n\in\mathbb{N}}\) is monotonically
decreasing and bounded from below, we consider the sequence \((-a_n)_{n\in\mathbb{N}}\), which is now
bounded from above and monotonically increasing. By the (already proven)
first statement of this theorem, the sequence \((-a_n)_{n\in\mathbb{N}}\) is convergent,
whence \((a_n)_{n\in\mathbb{N}}\) is
convergent as well.\(\Box\)
Remark 3. By the same argumentation as in Remark from page , the monotone increase (decrease) of \((a_n)_{n\in\mathbb{N}}\) can be slightly relaxed by only claiming that \(a_n\leq a_{n+1}\) (\(a_n\geq a_{n+1}\)) for all \(n\geq n_0\) for some \(n_0\) in \(\mathbb{N}\). In such a case, the limit of the sequence is then given by \(\sup\{a_n\,:\,n\geq n_0\}\) (resp. \(\inf\{a_n\,:\,n\geq n_0\}\)).
Example 4.
Consider the sequence \((a_n)_{n\in\mathbb{N}}\) which is recursively defined via \(a_1=1\) and \[a_{n+1}=\frac{a_n+\frac2{a_n}}{2}\;\text{ for $n\geq 1$.}\] We now prove that this sequence is convergent by showing that it is bounded from below and for all \(n\geq2\) holds \(a_{n+1}\leq a_n\).
Proof: To show boundedness from below, we use the inequality \(\sqrt{xy}\leq\frac{x+y}2\) for all nonnegative \(x,y\in\mathbb{R}\). This inequality is a consequence of \[0\leq\frac{(\sqrt{x}-\sqrt{y})^2}2=\frac{x+y}2-\sqrt{xy}.\] The first inequality is a consequence of the fact that squares of real numbers cannot be negative.
Using this inequality, we obtain for \(n\geq1\) \[a_{n+1}=\frac{a_n+\frac{2}{a_n}}{2}\geq \sqrt{a_n\cdot\frac2{a_n}}=\sqrt{2}.\] Thus, \((a_n)\) is bounded from below. For showing monotonicity, we consider \[a_{n+1}-a_n=\frac{a_n+\frac2{a_n}}{2}-a_n=\frac{1}{2a_n}(2-a_n^2).\] In particular, if \(n\geq2\), we have that \(a_n>0\) and \(2-a_n^2\leq0\). Thus, \(a_{n+1}-a_n\leq0\) for \(n\geq2\). An application of Theorem 2 (resp. the slight generalisation in Remark from above) now leads to the existence of some \(a\in\mathbb{R}\) with \(a=\lim_{n \to\infty}a_n\).
To compute the limit, we make use of the relation \(\lim_{n \to\infty}a_n=\lim_{n\to\infty}a_{n+1}\) (follows directly from the Definition of limits) and the formulae for limits. This yields \[a=\lim_{n \to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{a_n+\frac2{a_n}}{2}=\frac{a+\frac2{a}}{2}.\] This relation leads to the equation \(2-a^2=0\), i.e., we either have \(a=\sqrt{2}\) or \(a=-\sqrt{2}\). However, the latter solution cannot be a limit since all sequence elements are positive. Therefore, we have \[\lim_{n\to\infty}a_n=\sqrt{2}.\]Let \(x\in\mathbb{R}\) with \(x>1\). Consider the sequence \((\sqrt[n]{x})_{n\in\mathbb{N}}\). It can be directly seen that \((\sqrt[n]{x})_{n\in\mathbb{N}}\) is monotonically decreasing and bounded from below by one. Therefore, the limit \[a=\lim_{n\to\infty}\sqrt[n]{x}\] exists with \(a\geq1\). To show that \(a=1\), we assume that \(a>1\) and lead this to a contradiction.
The equation \(a>1\) leads to the existence of some \(n\in\mathbb{N}\) with \(a^n>x\), and thus \(a>\sqrt[n]{x}\). On the other hand, the monotone decrease of \((\sqrt[n]{x})_{n\in\mathbb{N}}\) implies that \[a=\lim_{n \to \infty}\sqrt[n]{x}=\inf\{\sqrt[n]{x}\,:\,n\in\mathbb{N}\}<a,\] which is a contradiction.Let \(x\in\mathbb{R}\) with \(0< x<1\). Consider the sequence \((a_n)_{n\in\mathbb{N}}=(\sqrt[n]{x})_{n\in\mathbb{N}}\). Then we have by Example b) and the rules for calculating limits that \[\lim_{n\to\infty}\sqrt[n]{x}=\frac1{\lim_{n\to\infty}\sqrt[n]{\frac1x}}=\frac11=1.\]
Let \((a_{n})\) be a nonnegative sequence with \(a_{n}\rightarrow a\) and \(k\in\mathbb{N}\). Then for all \(\varepsilon>0\) there exists \(N>0\) such that \(|a_{n}-a|<\varepsilon^{k}\). From this it follows that \[|\sqrt[k]{a_{n}}-\sqrt[k]{a}|\leq \sqrt[k]{|a_{n}-a|}<\varepsilon.\] Thus \((\sqrt[k]{a_{n}})\) is convergent with limit \(\sqrt[k]{a}\).
The sequence \((a_{n})_{n\in\mathbb{N}}\) defined as \(a_{n} := \left(1+\frac{1}{n}\right)^{n}\) is convergent.
Remark 5. The limit of the sequence \[(a_n)_{n \in \mathbb{N}N} = \left( \Big(1+\frac{1}{n} \,\Big)^{n}\right)_{n \in \mathbb{N}},\] i.e. \(e:=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}\) is well known as Euler’s number. Later on we will define the exponential function \(\exp\). It holds that \(e=\exp(1)\approx 2.7182818...\) . Indeed, we will show later on that \(e^{z}=\lim_{n\rightarrow\infty}\left(1+\frac{z}{n}\right)^{n}=\exp(z)\).
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