convergence of sequences

Section 2.1 convergence of sequences

Analysis is supposed to be about how functions behave. The most important thing a sequence can do (or not do) is converge.

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Definition 2.1.1.

We say that the sequence \(X=\left(x_n\right)_{n\in\mathbb{N}}\) converges to \(L\) if

for every \(\epsilon\gt 0\) there is \(K\in\mathbb{N}\) so that \(n\gt K\) guarantees \(\lvert x_n-L\rvert \lt \epsilon\text{.}\)
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in symbols,

\begin{equation*} \forall \epsilon\gt 0,\ \exists K\in\mathbb{N}:\ n\gt K\Rightarrow \lvert x_n-L\rvert \lt \epsilon \end{equation*}

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Convention 2.1.2.

We write

\begin{equation*} \lim X=L\ \ , \end{equation*}

\begin{equation*} X\to L\ \ , \end{equation*}

\begin{equation*} \lim x_n=L\ \ , \end{equation*}

\begin{equation*} x_n\to L\ \ , \end{equation*}

to express the statement that \(X=(x_n)_{n\in\mathbb{N}}\) converges to \(L\text{.}\)

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Example 2.1.3.

\begin{equation*} \frac{1}{n}\to 0 \end{equation*}

Proof.

Given \(\epsilon \gt 0\text{,}\) we know from Proposition 1.3.12 that there is some natural number \(K\) with \(\frac{1}{K}\lt \epsilon\text{.}\) We’ll use this \(K\text{.}\)

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To see that this \(K\) works, consider some \(n\gt K\text{.}\) We know that \(\frac{1}{n}\lt \frac{1}{K}.\)

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Therefore, \(n\gt K\) guarantees that

\begin{equation*} \lvert x_n-0\rvert=\lvert \frac{1}{n}\rvert=\frac{1}{n}\lt \frac{1}{K}\lt \epsilon \end{equation*}

which is what we need.

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The proof in Example 2.1.3 gives us a template for proving that a sequence converges. This template can be read off from the logical formula

\begin{equation*} \forall \epsilon\gt 0,\ \exists K\in\mathbb{N}:\ n\gt K\Rightarrow \lvert x_n-L\rvert \lt \epsilon \end{equation*}

as follows:

\(\forall \epsilon\) 🔗: We should start with "Given \(\epsilon\)".
\(\exists K\) 🔗: This tells us we need to propose a value for \(K\) and then verify that it does what we need.
\(n\gt K\Rightarrow \lvert x_n-L\rvert \lt \epsilon\) 🔗: There’s a hidden \(\forall n\) here, so we should start with "given \(n\gt K\)", and our goal will be to get from \(n\gt K\) to \(\lvert x_n-L\rvert\lt \epsilon\text{.}\)

You’ll find as we go through the semester that much of the work of introductory real analysis lies in setting up a proof outline from a logical sentence. If you’re not yet experienced in this kind of logical manipulation, don’t worry -- we’ll get plenty of practice. And that practice will certainly pay off.

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In a proof that sequence converges, the creative question is -- given \(\epsilon\text{,}\) how do we find \(K\text{?}\) In Example 2.1.3, \(K\) was handed to us on a silver platter by Proposition 1.3.12. But it’s not always so.

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Proposition 2.1.4.

Let \(X=\left(x_n\right)_{n\in\mathbb{N}}\) and \(Y=\left(y_n\right)_{n\in\mathbb{N}}\text{.}\) Consider the sequence \(X+Y=\left(x_n+y_n\right)_{n\in\mathbb{N}}\text{.}\) If \(X\to L\) and \(Y\to M\text{,}\) then \(X+Y\to L+M\text{.}\)

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Proof.

Given \(\epsilon\gt 0\text{,}\) consider the numbers \(\frac{1}{5}\epsilon\) and \(\frac{4}{5}\epsilon\text{.}\) Because \(\epsilon\gt 0\text{,}\) both \(\frac{1}{5}\epsilon \gt 0\) and \(\frac{4}{5}\epsilon\gt 0\text{.}\)

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Because \(X\to L\text{,}\) there is some \(I\in\mathbb{N}\) so that \(\lvert x_n-L\rvert \lt \frac{1}{5}\epsilon\text{.}\)

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Because \(Y\to M\text{,}\) there is some \(J\in\mathbb{N}\) so that \(\lvert y_n-M\rvert\lt \frac{4}{5}\epsilon\text{.}\)

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We’ll use \(K=I+J\text{.}\) To show that \(K\) works, let \(n\in\mathbb{N}\) have \(n\gt K\text{.}\) Notice that \(n\gt I\) and \(n\gt J\text{.}\) So

\begin{align*} \lvert (x_n+y_n)-(L+M)\rvert = \lvert (x_n-L)+(y_n-M)\rvert&\leq \lvert x_n-L\rvert +\lvert y_n-M\rvert\\ &\lt \frac{1}{5}\epsilon+\frac{4}{5}\epsilon = \epsilon \end{align*}

(provided that \(n\gt K\)).

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So this choice of \(K\) works.

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Remark 2.1.5.

You might wonder where the \(\frac{1}{5}\) and \(\frac{4}{5}\) came from. Why was that the right choice?

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Sorry to disappoint -- it wasn’t a particularly inspired choice. Most of the time I might have chosen \(\frac{1}{2}\) and \(\frac{1}{2}\text{.}\) All I needed was to make \(x_n\) very close to \(L\) and \(y_n\) very close to \(M\text{,}\) and arrange that those two "very closes" added up to \(\epsilon\) or less.

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Checkpoint 2.1.6.

Write out a version of the proof given in Proposition 2.1.4 which takes into account Remark 2.1.5 by using \(\frac{1}{100}\epsilon\) and \(\frac{1}{2}\epsilon\text{.}\)

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Of course we should also make sure we know what it means to not converge to some number, as in:

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Example 2.1.7.

\begin{equation*} 1+\frac{1}{n}\not\to 3 \end{equation*}

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Checkpoint 2.1.8.

State in words what it means for \(X\not\to L\text{.}\) Your answer should be as precise as Definition 2.1.1.

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State in logic what it means for \(X\not\to L\text{;}\) that is, what it means for

\begin{equation*} \forall \epsilon\gt 0,\ \exists K\in\mathbb{N}:\ n\gt K\Rightarrow \lvert x_n-L\rvert \lt \epsilon \end{equation*}

to be false.

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Theorem 2.1.9. Limits are unique.

If \(X\to L\) and \(X\to M\text{,}\) then \(L=M\text{.}\)

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Proof.

Fix some \(\epsilon\gt 0\text{.}\) We know that there is \(I\in\mathbb{N}\) so that \(n\gt I\) guarantees \(\lvert x_n-L\rvert\lt \frac{1}{2}\epsilon\) and some \(J\in\mathbb{N}\) so that \(n\gt J\) guarantees \(\lvert x_n-M\rvert\lt \frac{1}{2}\epsilon\text{.}\) So if we pick \(n\gt I+J\text{,}\) we’ll have both conditions. Picking such an \(n\text{,}\)

\begin{align*} \lvert L-M\rvert=\lvert L-x_n+x_n-M\rvert&\leq \lvert L-x_n\rvert+\lvert x_n-M\rvert \\ &=\lvert x_n-L\rvert+\lvert x_n-M\rvert\\ &\lt \frac{1}{2}\epsilon+\frac{1}{2}\epsilon=\epsilon \end{align*}

So we’ve shown that, no matter which positive number we pick, \(\lvert L-M\rvert\) is smaller. Since \(\lvert L-M\rvert\geq 0\text{,}\) it must be the case that \(\lvert L-M\rvert=0\text{.}\) Hence \(L=M\text{.}\)

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So we can refer to "the" limit of a sequence.

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Definition 2.1.10.

We call a sequence \(X=(x_n)_{n\in\mathbb{N}}\) convergent if it has a limit.

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In symbols,

\begin{equation*} \exists L:\ \forall \epsilon\gt 0,\ \exists K\in\mathbb{N}:\ n\gt K\Rightarrow \lvert x_n-L\rvert \lt \epsilon \end{equation*}

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Checkpoint 2.1.11.

What does it mean for a sequence to \(not\) be convergent? Give a symbolic version.

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Example 2.1.12.

The sequence \((-1)^n\) is not convergent.

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Definition 2.1.13.

The sequence \(X=(x_n)_{n\in\mathbb{N}}\) is bounded if its image is a bounded set. Equivalently, if there is a \(B\) so that \(\forall n, \lvert x_n\rvert\leq B\text{.}\)

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Theorem 2.1.14.

If \(X=(x_n)_{n\in\mathbb{N}}\) converges, then \(X\) is bounded.

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Proof.

\(X\) converges, say to the number \(L\in\mathbb{R}\text{.}\) Pick \(\epsilon=10\text{.}\) Then there is some \(K\in\mathbb{N}\) so that \(n\geq K\) guarantees \(\lvert x_n-L\rvert\lt 10\text{.}\) For such \(n\text{,}\)

\begin{equation*} \lvert x_n\rvert=\lvert x_n-L+L\rvert\leq \lvert x_n-L\rvert + \lvert L\rvert\lt 10+\lvert L\rvert \end{equation*}

so that every term past \(x_K\) has absolute value no greater than \(10+\lvert L\rvert\) (we say the \(K\)-tail of the sequence is bounded).

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Let

\begin{equation*} B=\max\left\{\lvert x_1\rvert,\lvert x_2,\cdots,\lvert x_K\rvert, L+10\right\} \end{equation*}

Then \(\forall n, \lvert x_n\rvert\leq B\text{.}\)

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Subsection 2.1.1 sequences and inequalities

An important question is which properties of a convergent sequence are inherited by its limit. For example, it’s quite easy to have a sequences of rational numbers which converges to an irrational, so ratonality is not inherited.

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The first such heritable property we’ll encounter is weak inequality:

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Theorem 2.1.15. limits preserve weak inequalities.

Let \(X=(x_n)_{n\in\mathbb{N}}\) be a convergent sequence: \(X\to L\text{.}\) If \(\forall n, x_n\leq A\text{,}\) then \(L\leq A\text{.}\)

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Proof.

If not, then \(L\gt A\text{.}\) So \(L-A\gt 0\text{.}\) Set \(\epsilon=\frac{1}{10}\left(L-A\right)\text{.}\)

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There is \(K\in\mathbb{N}\) so that \(n\geq K\) guarantees \(\lvert x_n-L\rvert\lt \frac{1}{10}(L-A)\text{.}\) Therefore,

\begin{equation*} x_K\geq L-\lvert x_K-L\rvert \gt L-\frac{1}{10}(L-A)=\frac{9}{10}L+A\gt \frac{9}{10}A+A=A \end{equation*}

which means that \(A\) wasn’t an upper bound for the sequence \(X\) after all.

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This contradiction shows that \(L\leq A\text{.}\)

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Checkpoint 2.1.16.

Prove the corresponding statement when \(\forall n\in\mathbb{N}, x_n\geq A\text{.}\)

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Remark 2.1.17.

It’s essential that the inequality in Theorem 2.1.15 is a weak inequality. Strong inequalities can be lost in the limit, as the sequence of positive numbers \(\frac{1}{n}\) shows.

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Sequences also give us a handle on suprema and infima.

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Proposition 2.1.18. sequential characterizations of supremum.

\(s=\sup A\) if and only if:

\(s\) is an upper bound for \(A\text{,}\) and

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there is a sequence \((a_n)_{n\in\mathbb{N}}\) so with \(\forall n,a_n\in A\) and \(a_n\to s\)

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\(s=\sup A\) if and only if:

there is a sequence \((B_n)_{n\in\mathbb{N}}\) so with \(\forall n,B_n\) is an upper bound for \(A\) and \(B_n\to s\)

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there is a sequence \((x_n)_{n\in\mathbb{N}}\) so with \(\forall n,x_n\) is not an upper bound for \(A\) and \(x_n\to s\)

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The sequential characterization of supremum ends up being pretty handy. Here’s a sequential proof that \(\sup(A+B)=\sup(A)+\sup(B)\text{:}\)

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Proof.

We’ll show that the clauses of the second version of Proposition 2.1.18 apply to \(\sup(A)+\sup(B)\text{.}\)

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Apply the second version separately to \(\sup A\) and \(\sup B\text{.}\) This gives the following sequences:

\((\alpha_n)_{n\in\mathbb{N}}\text{,}\) upper bounds for \(A\text{,}\) with \(\alpha_n\to \sup A\text{.}\)

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\((\beta_n)_{n\in\mathbb{N}}\text{,}\) upper bounds for \(B\text{,}\) with \(\beta_n\to \sup B\text{.}\)

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\((x_n)_{n\in\mathbb{N}}\text{,}\) non-upper bounds for \(A\text{,}\) with \(x_n\to\sup A\text{.}\)

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\((y_n)_{n\in\mathbb{N}}\text{,}\) non-upper bounds for \(B\text{,}\) with \(y_n\to\sup B\text{.}\)

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Now notice that \(\alpha_n+\beta_n\to \sup A+\sup B\) (using Proposition 2.1.4), and similarly \(x_n+y_n\to \sup A+\sup B\text{.}\) We need to show that each \(\alpha_n+\beta_n\) is an upper bound for \(A+B\text{,}\) and that each \(x_n+y_n\) is not an upper bound for \(A+B\text{.}\)

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Any \(z\in A+B\) has the form \(z=a+b\text{,}\) where \(a\leq \alpha_n\) and \(b\leq\beta_n\text{.}\) SO \(z\leq \alpha_n+\beta_n\) as required.

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For each \(n\text{,}\) there is \(a_n\in A\) and \(b_n\in B\) so that \(x_n\lt a_n\) and \(y_n\lt b_n\text{.}\) Thus \(x_n+y_n\lt a_n+b_n\text{,}\) which shows \(x_n+y_n\) is not an upper bound for \(A+B\text{.}\)

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Having constructed the two sequences \(\alpha_n+\beta_n\) and \(x_n+y_n\) which both converge to \(\sup A+\sup B\text{,}\) we’ve shown, using the other direction of Proposition 2.1.18, that \(\sup A+\sup B=\sup(A+B)\text{.}\)

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