Convergence
Convergent sequences have a well-defined limit.
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Study Convergence
Next we define the notions of convergence and limits:
Definition 1 (Convergence/divergence of sequences). Let \((a_n)_{n\in\mathbb{N}}\) be a sequence in \(\mathbb{K}\). We say that
\((a_n)_{n\in\mathbb{N}}\) is convergent to \(a\in \mathbb{K}\) if for all \(\varepsilon>0\) there exists some \(N=N(\varepsilon)\in\mathbb{N}\) such that for all \(n\geq N\) holds \(|a_n-a|<\varepsilon\). In this case, we write \[\lim_{n\to\infty}a_n=a.\]
\((a_n)_{n\in\mathbb{N}}\) is divergent if it is not convergent, i.e., for all \(a\in \mathbb{K}\) holds: There exists some \(\varepsilon>0\) such that for all \(N\) there exists some \(n>N\) with \(|a_n-a|\geq\varepsilon\).
Convergence for real sequences means that if you give any small distance \(\varepsilon\), one finds that all sequence members \(a_n\) lie in the interval \((a-\varepsilon, a+\varepsilon)\) with the exception of only finitely many.
Example 2.
Show: \((a_{n})_{n \in \mathbb{N}}\) with \(a_n = (1/n)\) is convergent with limit \(0\).
Show: \((b_{n})_{n \in \mathbb{N}}\) with \(b_n = (1/\sqrt{n})\) is convergent with limit \(0\).
Proof. Let \(\varepsilon > 0\). Choose \(N > \frac{1}{\varepsilon^2}\). Then for all \(n \geq N\), we have \[|b_n - 0| = \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{N}} < \varepsilon\] This means \(b_n\) is arbitrarily close to \(0\), eventually. ◻
Show: \((d_{n})_{n \in \mathbb{N}}\) with \(\displaystyle d_n = \frac{1}{\sqrt{1 + \frac{1}{n}}}\) is convergent with limit \(1\).
Proof. We know that \(\sqrt{1 + 1/n}\) is always greater than \(1\). Therefore, we can calculate: \[|1 - \frac{1}{\sqrt{1 + \frac{1}{n}}} |^2 \leq \frac{ 1- 2 \sqrt{1 + \frac{1}{n}} + 1 + \frac{1}{n}}{1 + \frac{1}{n}}\] \[\leq \frac{ 1- 2 + 1 + \frac{1}{n}}{1 + \frac{1}{n}} = \frac{ \frac{1}{n}}{1 + \frac{1}{n}} \leq \frac{1}{n}\] The square root is a monotonically increasing function, and hence, we can put the square root to both sides and get the inequality: \[0 ~\leq~ 1 - \frac{1}{\sqrt{1 + \frac{1}{n}}} ~\leq~ \frac{1}{\sqrt{n}}\] By the sandwich theorem of Homework 1, this shows that the sequence \((e_n)_{n \in \mathbb{N}N}\) given by \(e_n := 1 - \frac{1}{\sqrt{1 + \frac{1}{n}}}\) is convergent with limit \(0\). Then we use the limit theorem to get: \[\lim_{n \rightarrow \infty} d_n = \lim_{n \rightarrow \infty} (1 - e_n) = \lim_{n \rightarrow \infty} 1 - \lim_{n \rightarrow \infty} e_n = 1 - 0 = 1\] ◻
Remark 3.
It can be shown that for a complex sequence \((a_n)_{n\in\mathbb{N}}\), convergence to \(a\in\mathbb{C}\) holds true if, and only if, \((\Re(a_n))_{n\in\mathbb{N}}\) converges to \(\Re(a)\) and \((\Im(a_n))_{n\in\mathbb{N}}\) converges to \(\Im(a)\)
In fact, convergence can also be defined for sequences in some arbitrary normed vector space \(V\) (for the definition of a normed space, e.g. consult the linear algebra script). Then one has to replace the absolute value by the norm (e.g., “\(\|a_n-a\|<\varepsilon\)”.)
Due to the fact that for any \(N\in\mathbb{N}\), we can find some \(x\in\mathbb{R}\) with \(x>N\), we can equivalently reformulate the convergence definition as follows: “\((a_n)_{n\in\mathbb{N}}\) is convergent to \(a\in \mathbb{R}\) if for all \(\varepsilon>0\) there exists some \(N=N(\varepsilon)\in\mathbb{R}\) such that for all \(n\geq N\) holds \(|a_n-a|<\varepsilon\)”. In the following, we just write “there exists some \(N\)”.
What does convergence mean?
Outside any \(\varepsilon\)-neighbourhood of \(a\) only finitely many elements of the sequence exist.
Example 4.
The real sequence \((\frac1n)_{n\in\mathbb{N}}\) converges to 0.
Proof: Let \(\varepsilon>0\) (be arbitrary): Choose \(N=\frac1\varepsilon+1=\frac{1+\varepsilon}\varepsilon\).
Then for all \(n\geq N\) holds \[\frac1n\leq\frac1N=\frac{\varepsilon}{1+\varepsilon}<\varepsilon.\] Therefore \[\left|\frac1n-0\right|=\frac1n<\varepsilon.\]The real sequence \(((-1)^n)_{n\in\mathbb{N}}\) is divergent.
Proof by contradiction:
Assume that \(((-1)^n)_{n\in\mathbb{N}}\) is convergent to \(a\in\mathbb{R}\). Then take e.g. \(\varepsilon=\frac1{10}\). Due to convergence, we should have some \(N\) such that for all \(n\geq N\) holds \[|(-1)^n-a|<{\textstyle\frac1{10}}.\] Thus we have that \(|-1-a|<\frac1{10}\) and \(|1-a|<\frac1{10}\). As a consequence, \[2=|1+a-a+1|\leq|1+a|+|-a+1|=|1+a|+|a-1|<{\textstyle\frac1{10}+\frac1{10}=\frac1{5}}.\] This is a contradiction.\(\Box\)For \(q\in\mathbb{C}\backslash\{0\}\) with \(|q|<1\) the complex sequence \((q^n)_{n\in\mathbb{N}}\) converges to 0.
Proof: \(|q|<1\) gives rise to \(\frac1{|q|}>1\), whence \(\frac1{|q|}-1>0\). Therefore, we are able to apply Bernoulli’s inequality (see tutorial) in the following way:\[\frac1{|q|^n}=\left(1+\left(\frac1{|q|}-1\right)\right)^n =\left(1+\left(\frac{1-|q|}{|q|}\right)\right)^n\geq 1+n\cdot \left(\frac{1-|q|}{|q|}\right),\] and thus \[|q|^n\leq \frac1{1+n\cdot \left(\frac{1-|q|}{|q|}\right)}=\frac{|q|}{|q|+n\cdot ({1-|q|})}.\] Now let \(\varepsilon>0\) (be arbitrary):
Choose \[N=\frac{|q|}{\varepsilon\cdot(1-|q|)}-\frac{|q|}{1-|q|}+1\] Then for all \(n\geq N\) holds \[n>\frac{|q|}{\varepsilon\cdot(1-|q|)}-\frac{|q|}{1-|q|}\] and thus \[n\cdot ({1-|q|})>\frac{|q|}{\varepsilon}-|q|.\] This leads to \[|q|+n\cdot ({1-|q|})>\frac{|q|}{\varepsilon},\] whence \[\frac{|q|}{|q|+n\cdot ({1-|q|})}<\varepsilon.\] The above calculations now imply \[|q^n-0|=|q|^n\leq\frac{|q|}{|q|+n\cdot ({1-|q|})}<\varepsilon.\] \(\Box\)
Remark 5. The choice of the \(N\) often seems “to appear from nowhere”. However, there is a systematic way to formulate the proof. For instance in a), we need to end up with the equation \(|\frac1n-0|<\varepsilon\) or, equivalently, \(\frac1n<\varepsilon\). Inverting this expression leads to \(n>\frac1\varepsilon\). Therefore, if \(N\) is chosen as \(N=\frac1\varepsilon+1=\frac{1+\varepsilon}\varepsilon\), the desired statement follows.
If one has to formulate such a proof (for instance, in some exercise), then first these above calculations have to be done “on some extra sheet” and then formulate the convergence proof in the style as in a) or c).
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