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2.5 Increasing and decreasing sequences

Definition 2.5.1

A sequence $\{x_{n}\}^{\infty}_{n=1}$ is increasing if for any $n\geq 1$ $x_{n}\leq x_{n+1}$ . The sequence is decreasing if for any $n\geq 1$ $x_{n}\geq x_{n+1}$ .

Example 2.5.1

The following examples are crucial:

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The sequence $1,2,3,4\dots$ is an increasing sequence.
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The constant sequence $1,1,1,1,1,\dots$ is still an increasing sequence.
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$\{x_{n}=\frac{n}{n+1}\}^{\infty}_{n=1}$ is an increasing sequence.
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Let $\{y_{n}\}^{\infty}_{n=1}$ be a non-negative sequence of real numbers. Then $y_{1},y_{1}+y_{2},y_{1}+y_{2}+y_{3},\dots$ is an increasing sequence.
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If the sequence $\{x_{n}\}^{\infty}_{n=1}$ is increasing, then the sequence $\{-x_{n}\}^{\infty}_{n=1}$ is decreasing.

Theorem 2.5.1

Every bounded increasing (decreasing) sequence $\{x_{n}\}^{\infty}_{n=1}$ is convergent.

Proof: Let $y$ be the least upper bound of the sequence $\{x_{n}\}^{\infty}_{n=1}$ . We know that for any $\varepsilon>0$ there exists $x_{n_{\varepsilon}}$ so that $y-x_{n_{\varepsilon}}<\varepsilon$ . Therefore, $x_{n}\to y$ , since for any $\varepsilon>0$ if $n>n_{\varepsilon}$ , then $|y-x_{n}|<\varepsilon$ . $\square$

The following theorem is somewhat stronger than the Bolzano-Weierstrass Theorem, so by proving it we will obtain yet another proof of the BWT.

Theorem 2.5.2

Let $\{x_{n}\}^{\infty}_{n=1}$ be a convergent sequence, then it contains either a decreasing or an increasing subsequence.

Proof: The proof goes like a computer program. Let $y_{1}$ be the least upper bound (supremum) of the sequence $\{x_{n}\}^{\infty}_{n=1}$ . If $y_{1}$ is not a maximum, so $y_{1}>x_{n}$ for all $n\geq 1$ , then we must have a sequence $x_{m_{1}}<x_{m_{2}}<\dots$ increasing sequence and we can STOP the proof. If $y_{1}=x_{n_{1}}$ , then we consider the sequence $x_{n_{1}},x_{n_{1}+1},x_{n_{1}+2},\dots$ . Let $y_{2}$ be the supremum of this sequence. If $y_{2}$ is not a maximum, we can again STOP the proof. If it is a maximum we can GO TO the next step, $y_{2}=x_{n_{2}},y_{1}\geq y_{2}$ and consider the sequence $x_{n_{2}},x_{n_{2}+1},x_{n_{2}+2},\dots$ . This program either stops, that is find an increasing sequence, or it never stops and finds the decreasing sequence $y_{1}\geq y_{2}\geq y_{3}\geq\dots$ . $\square$

Example 2.5.2

It is possible that a convergent sequence contains both increasing and decreasing subsequences. $0,2,1/2,3/2,3/4,4/3,4/5,5/4,\dots$ will converge to $1$ . The subsequence $0,1/2,2/3,3/4,4/5,\dots$ is increasing, the subsequence $2,3/2,4/3,5/4,\dots$ is decreasing.