(last updated: 5:11:25 AM, October 09, 2020)
Now we will be introduced to the idea of convergence of sequences. This is a concept you have already seen, probably in Calculus I & II, but here we will develop it rigorously meaning it will follow precisely from the axiomatic structure of the real number system. We define very specifically what it precisely means for a sequence to converge.
Definition 2.1.1 A sequence is a real-valued function whose domain is a set of the form \(\{n\in\mathbb Z \mid n\geq m\}\), when \(m\) is a fixed integer.
Typically \(m=1\). For example \(\{1/n\}\) is the sequence \(1,\frac12,\frac13,\ldots\) (assuming \(n=1,2,\ldots\)).
In function notation, we’ll see a formula like \(a_n=f(n)\) such as \(a_n=\frac1n\) (i.e. \(f(n)=\frac1n\)) with the subscript \(n\) called the index and the \(a_n\) evaluated at each \(n\) called the terms. It is often written as \[f:\mathbb N\rightarrow \mathbb R\] to denote that \(f\) is a function with domain \(\mathbb N\) and codomain \(\mathbb R\). Having \(\mathbb R\) as its codomain (distinct from the range, technically) tells us it is a real-valued function. Having \(\mathbb N\) as the domain tells us that it is a sequence, a discrete list of numbers with a starting point that typically has no end.
Also recall that a function \(f: S\rightarrow f(S)\) is a relation and is a subset of the cartesian product \(S\times f(S)\), \(f\subset S\times f(S)\). We have that \(f=\{(x,f(x)) \mid x\in Dom(f)\}\). Since a sequence is a function, \(a_n=f(n)\) with \(f:\mathbb N\rightarrow R\), then we can say that a sequence is really the set \(\{(n,a_n) \mid n\in\mathbb N\}\). So when we write \(a_1,a_2,\ldots\), we really mean: \[(1,a_1), (2,a_2),\ldots, (k,a_{k}), (k+1,a_{k+1}), \ldots\]
Note on notation: Often a sequence will be denoted by subtly different notations including: \(x_n\), \(\{x_n\}\), \(\{x_n\}_{n\in\mathbb N}\), \(\{x_n\}_{n=1}^\infty\). These are all generally interchangeable. You will see different conventions in different textbooks and by different authors. Another common notation is \((x_n)\) or \((x_n)_{n\in\mathbb N}\). You have to practice picking out the meaning of the notation from the context: \(x_n\) could refer to a specific value of the sequence at index \(n\), or it could just represent the sequence generally with \(n\) ranging over the natural numbers.
Negative indexes are also allowed, e.g. \(\left\{\frac{k+7}{k^2+1}\right\}_{k\geq-5}\) giving the sequence \(\frac{2}{26},\frac{3}{17},\frac{4}{10},\ldots\).
Definition 2.1.2 A sequence \(\{a_n\}\) converges to a real number \(A\) if and only if for each real number \(\epsilon>0\), there exists a positive integer \(n^*\) such that \[|a_n-A|<\epsilon \quad \text{ for all } \quad n\geq n^*.\]
You can normally think of \(\epsilon\) as a very small positive number like \(\epsilon=\frac1{100}\). A smaller \(\epsilon\) means we want to show our sequence can get closer to the limiting value \(A\). Usually, a sequence can be shown to be convergent once we find how the cut-off \(n^*\) could depend on \(\epsilon\). It is usually the case that a smaller \(\epsilon\) (getting closer to the limiting value \(A\)) requires a larger cut-off \(n^*\) value.
We say that “\(a_n\) converges to \(A\) (as \(n\) goes to infinity)”. We can write it in a few different ways such as \[a_n\rightarrow A \ \text{ as } \ n\rightarrow\infty\] or one of my preferred notations is \[a_n \overset{n\rightarrow\infty}{\longrightarrow} A\] Often we just write \(a_n\rightarrow A\) as it is usually to be understood that \(n\) is ranging over the natural numbers.
Example: Consider the sequence \(a_n=\frac{n}{n^2+1}\). You might intuitively understand that this sequence coverges to zero as \(n\) goes to infinity. But we need to show this with a careful/rigorous mathematical argument.
Scratch work: We want to find a cut-off \(n\)-value \(n^*\) such that when \(n\geq n^*\), \(|a_n-0|<\epsilon\). This gives \(\frac{n}{n^2+1}<\epsilon\). Can we rearrange this to find how \(n\) depends on \(\epsilon\)?
Consider the calculation: \[ \begin{aligned} \frac{n}{n^2+1}&<\frac{n}{n^2} \quad & {\small \text{(smaller denominator $\Rightarrow$ bigger fraction)}}\\ &=\frac{1}{n} \quad &{\small \text{(algebra)}}\\ &<\epsilon \quad &{\small \text{(what we want)}} \end{aligned} \] It happens that \(\frac{1}{n}<\epsilon\) will occur when \(\frac{1}{\epsilon}<n\). So when we are given any \(\epsilon>0\), all we have to do is choose any \(n^* > \frac{1}{\epsilon}\) and that gives \(|a_n-0|<\epsilon\). Notice that \(n^* > \frac{1}{\epsilon}\) also rearranges to \(\frac1{n^*} < \frac1{1/\epsilon}\). Now we write up our formal argument:
Formal argument: (this is what you would want to write and turn in) Let \(\epsilon>0\) and chose \(n^*>\frac1\epsilon\). Then when \(n>n^*\) we have that \[ \begin{aligned} \left|\frac{n}{n^2+1}-0\right|&=\frac{n}{n^2+1}\\ &<\frac{n}{n^2} \\ &=\frac{1}{n} \\ &\leq\frac{1}{n^*}\\ &<\frac{1}{1/\epsilon} \quad {\small\text{(since $n^*>1/\epsilon$)}}\\ &=\epsilon \end{aligned} \] So we have that \(\frac{n}{n^2+1} < \epsilon\) when \(n \geq n^*>\frac1\epsilon\). Therefore \(\frac{n}{n^2+1}\rightarrow 0\) as \(n\rightarrow\infty\). \(\square\)
Example: Show that \(a_n=\frac{n^2-10}{n^2-20}\) converges to \(1\).
We want \(\left| \frac{n^2-10}{n^2-20}-1 \right|<\epsilon\). First notice that no \(n\in\mathbb N\) gives a zero in the denominator. Now simplify: \(\left| \frac{n^2-10}{n^2-20}-1 \right|=\left| \frac{10}{n^2-20}\right|\) so that we need to show \(\left| \frac{10}{n^2-20}\right|<\epsilon\). This is tricky because of the subtraction in the denominator! As long as \(n\geq 5\) we have \(\left| \frac{10}{n^2-20}\right|= \frac{10}{n^2-20}\) though. Now we just need to solve for \(n\) as a function of \(\epsilon\): \(\frac{10}{n^2-20}<\epsilon\) gives \(\sqrt{\frac{10}\epsilon+20}<n\). So as long as we choose a cut-off value \(n^*>\sqrt{\frac{10}\epsilon+20}\) then the argument will work… Almost!
Remember that this calculation originally needed \(n\geq5\) to keep the denominator positive. So we need to satisfy \(n^*\geq5\) as well, and simultaneously satisfy \(n^*>\sqrt{\frac{10}\epsilon+20}\) so that for any \(n\geq n^*\) we have \(n\geq5\) AND \(n>\sqrt{\frac{10}\epsilon+20}\). We solve this by choosing some \[n^*>\max\left\{5,\sqrt{\frac{10}{\epsilon}+20}\right\}.\]
Formal write-up: Let \(\epsilon>0\) and choose some \(n^*>\max\left\{5,\sqrt{\frac{10}{\epsilon}+20}\right\}\).
Then \[ \begin{aligned} \left| \frac{n^2-10}{n^2-20}-1 \right|&=\left| \frac{10}{n^2-20}\right|&\\ &=\frac{10}{n^2-20}\quad &{\small\text{(since $n \geq n^*>5$)}}\\\\ &<\frac{10}{\left(\sqrt{\frac{10}{\epsilon}+20}\right)^2-20}\quad & \left({\small\text{since $n\geq n^*>\sqrt{\frac{10}{\epsilon}+20}$}}\ \right)\\ &=\epsilon& \end{aligned} \] Thus \(\frac{n^2-10}{n^2-20}\) converges to \(1\). \(\square\)
Note that the inequalities in the definition of convergence are somewhat flexible.
\[|a_n-A|<\epsilon \quad \text{ for all } \quad n\geq n^*\] \[|a_n-A|\leq\epsilon \quad \text{ for all } \quad n> n^*\] \[|a_n-A|\leq\epsilon \quad \text{ for all } \quad n \geq n^*\] \[|a_n-A|<\epsilon \quad \text{ for all } \quad n> n^*\] Are all equivalent for the purposes of determining convergence. We just need to always have freedom to choose a smaller \(\epsilon\) and make the sequence closer to the limiting value. So you do not need to worry too much above which version you have shown. It is a bit mathematically “cleaner” to show “\(<\epsilon\)” as opposed to “\(\leq\epsilon\)” though.
The next theorem proof contains a very common and useful technique that I call the \(\epsilon/2\) technique. If we want to show that \(|A+B|<\epsilon\) but we only know how to show \(|A|<\epsilon\) and \(|B|<\epsilon\), then by the triangle inequality we have \(|A+B|\leq|A|+|B|\) so if we can get \(|A|<\epsilon/2\) and \(|B|<\epsilon/2\) then our desired result follows.
Definition 2.1.6 If for all real numbers \(A\), \(a_n\) does not converge to \(A\), then \(a_n\) diverges.
Example: Prove that the sequence \((-1)^n\) diverges.
Let \(\epsilon=\frac12\) and assume that \((-1)^n\) converges to some real number \(A\). Then there is an \(n^*\) such that \(n\geq n^*\) implies that \(|(-1)^n-A|<\frac12\). Now when \(n\) is even or odd, this will give use two different equations, and we know that beyond any \(n^*\), there are infinitely many even and odd natural numbers. \[|(-1)^n-A|= \begin{cases} |-1-A| & n \text{ is odd}\\ |1-A| & n \text{ is even}\\ \end{cases} \] Thus \(|-1-A|=|1+A|<\frac12\) and \(|1-A|<\frac12\). But these inequalities give us that \(-\frac12<1+A<\frac12\) and \(-\frac12<1-A<\frac12\) must simultaneously be true. Rearranging these gives \(-1-\frac12<A<-1+\frac12\) and \(1-\frac12<A<1+\frac12\) (this last one takes careful rearranging). In other words, \(A\in(-1.5,-0.5)\) and \(A\in(0.5,1.5)\), but these are disjoint intervals. Thus we have a contradiction since \(A\) cannot simultaneously be in both intervals. \(\square\)
\(\bullet\) How is this statement different from the previous one?
Theorem 2.1.9 Any two limits of a convergent sequence are the same.
Proof. Assume that \(a_n\rightarrow A\) and \(a_n\rightarrow B\) (both as \(n\rightarrow\infty\)). Then we will show that \(A=B\) must follow.
Let \(\epsilon>0\) and choose \(n_1\) such that \(|a_n-A|<\epsilon/2\) for any \(n>n_1\). We can do this since \(\epsilon/2>0\) thus we can always choose a large enough index byond which the sequence is within \(\epsilon/2\) of its limit. Similarly, choose \(n_2\) sich that \(|a_n-B|<\epsilon/2\) for any \(n>n_2\).
Now let \(n^*>\max\{n_1,n_2\}\) so that \(n^*>n_1\) and \(n^*>n_2\). Then for all \(n\geq n^*\) \[ \begin{aligned} |A-B|&=|A-a_n+a_n-B|\\ & \leq |A-a_n|+|a_n-B|\\ &<\frac{\epsilon}{2}+\frac\epsilon2\\ &=\epsilon \end{aligned} \] Thus for any \(\epsilon>0\), we have shown that \(|A-B|<\epsilon\). This implies that \(|A-B|=0\), i.e. that \(A=B\). \(\blacksquare\)
Definition 2.1.10 A sequence \(\{a_n\}\) is said to be bounded if and only if there exists a positive constant \(M\) such that \(|a_n|\leq M\) for all \(n\in\mathbb N\). If a sequence is not bounded, then it is said to be unbounded.
A sequence can also be bounded from above or bounded from below. A bounded sequence is bounded from above and below.
Theorem 2.1.11 Any convergent sequence is bounded.
Proof. Suppose \(a_n\) converges to \(A\). This means that for any \(\epsilon>0\), there is an \(n^*\) such that for any \(n\geq n^*\), \(|a_n-A|<\epsilon\).
In particular, let \(\epsilon=1\), then for some \(n^*\), \(|a_n-A|<1\) when \(n\geq n^*\). This means that \(-1<a_n-A<1\) for all sufficiently large indexes \(n\). Then we can calculate that \[ \begin{aligned} |a_n|&=|a_n-A+A|&\\ &\leq|a_n-A|+|A| \quad &\text{(by triangle ineq.)}\\ &<1+|A| \quad &\text{(since $|a_n-A|<1$)} \end{aligned} \] So the “tail” of the sequence is definitely bounded, that \(|a_n|<1+|A|\) for \(n=n^*,n^*+1,\ldots\).
Now \(\{a_1,a_2,\ldots,a_{n^*-1}\}\) is a finite set of real numbers. Let \(K=\max_{n<n^*}\{|a_n|\}\). Then \(|a_n|\leq K\) for all \(n<n^*\). Note that we are taking the maximum of the absolute value of the sequence terms. This is important.
Now let \(M=\max\{|A|+1,K\}\), and this gives us that \(|a_n|\leq M\) for all \(n\in\mathbb N\). Thus the sequence \(\{a_n\}\) is bounded. \(\blacksquare\)
Theorem 2.1.12 Consider a sequence \(\{a_n\}\) that converges to a nonzero constant \(A\). Then there exists \(n^*\) such that \(a_n\neq0\) for all \(n\geq n^*\). In fact, (there exists and \(n^*\) such that) \(a_n\geq\frac12|A|\) for all \(n\geq n^*\).
Proof. \(A\neq0\) therefore \(|A|/2>0\). Let \(\epsilon=\frac12|A|\). Then we know that there is an \(n^*\) such that for all \(n\geq n^*\) \(|a_n-A|<\epsilon=\frac12|A|\). This means that \(-\frac12|A|<a_n-A<\frac12|A|\), and adding \(|A|\) across the inequality gives \(|A|-\frac12|A|<a_n<|A|+\frac12|A|\). This gives \(\frac12|A|<a_n\) (for any \(n\) greater than or equal to our \(n^*\)). \(\blacksquare\)
Theorem 2.1.13: If a real number, \(r\), satisfies \(|r|<1\), then \(\displaystyle\lim_{n\rightarrow\infty} r^n = 0\).
Theorem 2.1.13 is extremely important and will be useful in many contexts. I will not present the proof, but you should read the proof in the text though. It uses the binomial theorem, and it is a nice proof. This theorem is much easier to prove once we know more about increasing and decreasing bounded sequences which are covered in Section 2.4.
Example: Let \(k\in\mathbb N\), then \(\frac{k}{k+1}<1\). So we have that the sequence \(a_n=\left(\frac{k}{k+1}\right)^n\) converges to zero.
Example: Let \(m,k\in\mathbb N\), then \(\frac{k}{k+m}\leq\frac{k}{k+1}<1\). So we have that the sequence \(b_n=\left(\frac{k}{k+n}\right)^n\) converges to zero since \(0<b_n=\left(\frac{k}{k+n}\right)^n\leq\left(\frac{k}{k+1}\right)^n\). For any \(\epsilon>0\) we can choose a cut-off \(n^*\) such that \(0<b_n\leq\left(\frac{k}{k+1}\right)^n<\epsilon\) for all \(n\geq n^*\). Thus \(|b_n-0|<\epsilon\) for all \(n\geq n^*\), hence \(b_n\) converges to zero.
Here is a Remark that is not in your textbook, but these inequalities can help you simplify equations and find nicer bounds.
Remark: (not in text) Notice that if \(0<r<1\) then \[0<r^2<r<\sqrt r<1\] and that if \(1<r\), then \[1<\sqrt r<r<r^2\]
We can similarly argue that if \(0<r<1\), then \[0<r^{n+1}<r^{n}<r<r^{1/n}<r^{1/(n+1)}<1\] and that if \(1<r\), then \[1<r^{1/(n+1)}<r^{1/n}<r<r^{n}<r^{n+1}\] for all \(n\in\mathbb N\), \(n\geq2\).
Also, let \(n\in\mathbb N\). We can raise \(n\) to powers and take roots and we have that \(n^k\leq n^{k+1}\) and \(\sqrt[k+1]n\leq \sqrt[k]n\) for all \(k\in\mathbb N\). These inequalities are all strict \(<\) if \(n>1\).
Keep the above remark in mind as it will be useful qite often.
Example: Let \(a_n=\sqrt{1-\frac1n}\) for \(n\in\mathbb N\). If we want to prove that \(a_n\rightarrow 1\), then it might be nice to show that \(0<1-\frac1n<1\), thus \(0<1-\frac1n<\sqrt{1-\frac1n}<1\). Now \[ \begin{aligned} \left|\sqrt{1-\frac1n}-1\right|&=1-\sqrt{1-\frac1n}\quad&\left(\text{since $0<\sqrt{1-\frac1n}<1$}\right)\\ &<1-(1-\frac1n)\quad&\left(\text{since $0<1-\frac1n<\sqrt{1-\frac1n}<1$}\right)\\ &=\frac1n.& \end{aligned} \] We know that for any \(\epsilon>0\) we can choose \(n^*>\frac1\epsilon\) such that \(n\geq n^*\) implies \(\frac1n<\epsilon\) and hence this gives us the convergence of \(a_n=\sqrt{1-\frac1n}\) that we seek.
Theorem 2.1.14: The sequence \(\{a_n\}\) converges to \(0\) if and only if the sequence \(\{|a_n|\}\) converges to \(0\).
Proof. Note that \(|a_n|=|a_n-0|=\big| |a_n|-0\big|\). Thus \(|a_n-0|<\epsilon\) if and only if \(\big| |a_n|-0\big|<\epsilon\). This is essentially enough of an argument, but here it is more clearly written out:
If \(a_n\) converges to zero, then for any \(\epsilon>0\) we can find an \(n^*\in\mathbb N\) such that for any \(n\geq n^*\) we have that \(\epsilon>|a_n-0|=\big| |a_n|-0\big|\), and hence \(|a_n|\) converges to zero.
If \(|a_n|\) converges to zero, then for any \(\epsilon>0\) we can find an \(n^*\in\mathbb N\) such that for any \(n\geq n^*\) we have that \(\epsilon>\big| |a_n|-0\big|=|a_n-0|\), and hence \(a_n\) converges to zero. \(\blacksquare\)
Note that the above theorem does not work if the limit is not zero! Example: \(a_n=(-1)^n\) does not converge, but \(|a_n|\) converges to one.
Definition 2.1.15 A property, \(P\), is true eventually if and only if there exists \(M > 0\) such that \(P\) is true for all \(x \geq M\).
Example:
\(\frac1n\) is eventually less than \(\epsilon=0.00001\).
Given \(f(x)=x^2\) for \(x\in\mathbb R\) and any \(n\in\mathbb N\). Eventually, \(f(x)>n\).
If \(a_n\) converges to \(A\), then, given any \(\epsilon>0\), \(a_n\) is eventually less than a distance of \(\epsilon\) from \(A\).
\[ \diamond \S \diamond \]