26 Cauchy Sequences Practice

#todo/school: Finish 2.6.2, 2.6.4, 2.6.5 📅 2024-10-27 ✅ 2024-10-28

2.6.2

Question

Give an example of each of the following, or argue that such a request is impossible.
a. A Cauchy sequence that is not monotone.
b. A Cauchy sequence with an unbounded subsequence.
c. A divergent monotone sequence with a Cauchy subsequence.
d. An unbounded sequence containing a subsequence that is Cauchy.

Proof

a. The sequence:

(a_{n}) = {\begin{cases} \frac{- 1}{n} & x is even \\ \frac{1}{n} & x is odd \end{cases}

That's because the sequence isn't strictly increasing or decreasing (look at the first few terms to see the alternating behavior. But it is convergent to 0 and thus is Cauchy.

b. Impossible. From 25 Subsequences and the Bolzano-Weierstrass Theorem Practice#2.5.5 the contra-positive says that if there is a subsequence that diverges, then the main (bounded) subsequence diverges. If the Cauchy sequence (which must be bounded since it's convergent) had an unbounded subsequence, then the Cauchy sequence must diverge which is impossible.

c. Impossible. A divergent monotone sequence is unbounded (assume that the sequence is bounded. Then MCT says is convergent which is a contradiction), so then any subsequence is still monotone and unbounded. That implies that all these sub-sequences are divergent as the contra-positive of MCT.

d. Take:

(a_{n}) = {\begin{cases} n & x is odd \\ \frac{1}{n} & x is even \end{cases}

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2.6.3

Question

If $(x_{n})$ and $(y_{n})$ are Cauchy sequence, then one easy way to prove that $(x_{n} + y_{n})$ is Cauchy is to use the Cauchy Criterion. By Cauchy Sequences Converge, $(x_{n})$ and $(y_{n})$ converge and the Limit Laws (Algebraic Limit Theorem) implies $(x_{n} + y_{n})$ is convergent, hence Cauchy.
a. Give a direct argument that $(x_{n} + y_{n})$ is a Cauchy sequence that does not use the Cauchy Criterion or the ALT.
b. Do the same for the product $(x_{n} y_{n})$ .

Proof

a. Let $ϵ > 0$ . Notice $\frac{ϵ}{2} > 0$ then because $x_{n}, y_{n}$ are Cauchy (since they're convergent) then there is some $N_{x}, N_{y} \in N$ such that for all $m, n \geq N_{x}$ :

| x_{n} - x_{m} | < ϵ

and for $m, n \geq N_{y}$ :

| y_{n} - y_{m} | < ϵ

Now take $N = max (N_{x}, N_{y})$ . Then let any $m, n \geq N$ so then:

\begin{aligned} | (x + y)_{n} - (x + y)_{m} | & = | (x_{n} - x_{m}) + (y_{n} - y_{m}) | \\ \leq | x_{n} - x_{m} | + | y_{n} - y_{m} | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} \\ = ϵ \end{aligned}

b. Bound $| x_{n} | \leq M_{1}$ and $| y_{n} | \leq M_{2}$ are bounds (since they converge), then choose $N = max (N_{x}, N_{y})$ (same as (a)) such that:

\begin{aligned} | x_{n} y_{n} - x_{m} y_{m} | & = | (x_{n} y_{n} - x_{n} y_{m}) - (x_{n} y_{m} - x_{m} y_{m}) | \\ \leq | x_{n} (y_{n} - y_{m}) | + | y_{m} (x_{n} - x_{m}) | \\ = | x_{n} | | y_{n} - y_{m} | + | y_{m} | | x_{n} - x_{m} | \\ \leq M_{1} | y_{n} - y_{m} | + M_{2} | x_{n} - x_{m} | \\ < M_{1} (\frac{ϵ}{2 M_{1}}) + M_{2} (\frac{ϵ}{2 M_{2}}) \\ = ϵ \end{aligned}

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2.6.4

Question

Let $(a_{n})$ and $(b_{n})$ be Cauchy sequences. Decide whether each of the following sequence is a Cauchy sequence, justifying each conclusion.
a. $c_{n} = | a_{n} - b_{n} |$
b. $c_{n} = (- 1)^{n} a_{n}$
c. $c_{n} = [[a_{n} ∥ a_{n}]]$ where $[[x ∥ x]]$ refers to the greatest integer less than or equal to $x$ .

Proof

a. Yes it's Cauchy since when $(a_{n}), (b_{n})$ are Cauchy you can use 26 Cauchy Sequences Practice#2.6.3 (a) to get that their sum must also be Cauchy (the absolute values don't change anything here, since we know $(a_{n} - b_{n})$ must be Cauchy, the sequence is definitely convergent for the absolute values, applying the definition).

b. No. For example if $a_{n} = 1$ then $c_{n} = (- 1)^{n}$ which doesn't converge and thus can't be Cauchy.

c. Not it's possible for it to not be Cauchy. Take:

a_{n} = \frac{(- 1)^{n}}{n}

this doesn't work since for the odd $a_{n}$ 's the $[[a_{2 k + 1} ∥ a_{2 k + 1}]] \to - 1$ while $[[a_{2 k} ∥ a_{2 k}]] \to 0$ which are different values, so then by the contrapositive of 25 Subsequences and the Bolzano-Weierstrass Theorem Practice#2.5.5 then this must diverge and thus not be Cauchy.
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2.6.5

Question

Consider the following (invented) definition: A Sequence $s_{n}$ is pseudo-Cauchy if, $\forall ϵ > 0$ there $\exists N \in N$ such any $n \geq N$ implies $| s_{n + 1} - s_{n} | < ϵ$ .
Decide which one of the following two propositions is actually true. Supply a proof for the valid statement and a counterexample for the other.
i. Pseudo-Cauchy sequences are bounded.
ii.. If $(x_{n}), (y_{n})$ are pseudo-Cauchy, then $(x_{n} + y_{n})$ is pseudo-Cauchy as well.

Scratch:

Let's see if (i) is true

The idea of (i) is that any step $ϵ$ has all terms after some $N$ taking that step size between values. That would make sequences like $\frac{1}{n}$ are still pseudo-Cauchy and thus convergent.

What about (ii)? It's probably not true since:

\begin{aligned} | (x_{n + 1} + y_{n + 1}) - (x_{n} + y_{n}) | & \leq | x_{n + 1} - x_{n} | + | y_{n + 1} - y_{n} | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} \\ = ϵ \end{aligned}

So this property does hold for (ii), so (i) must not be true. Essentially, pseudo-Cauchy is weaker than convergence (being fully Cauchy).

Proof

i. This is false. Consider the sequence $s_{n} = \sum_{i}^{n} \frac{1}{i}$ . Then notice that $s_{n}$ is unbounded, but it is pseudo-Cauchy as we'll show. Let $ϵ > 0$ . We want to show that:

| s_{n + 1} - s_{n} | < ϵ

The LHS reduces to:

| \sum_{i = 1}^{n + 1} \frac{1}{i} - \sum_{i = 1}^{n} \frac{1}{i} | = \frac{1}{n + 1}

Notice that for any $ϵ$ if we have:

\begin{aligned} \frac{1}{n + 1} & < ϵ \\ \frac{1}{ϵ} & < n + 1 \\ \frac{1}{ϵ} - 1 & < n \end{aligned}

we need to choose $N > \frac{1}{ϵ} - 1$ . Then any $n > N$ (working backwards) has:

| s_{n + 1} - s_{n} | < ϵ

showing pseudo-Cauchy for $s_{n}$ . However, as we know $s_{n}$ is unbounded (namely because it's divergent and increasing) so this is a counter-example to (i).

ii. This is true. Let $(x_{n}), (y_{n})$ be pseudo-Cauchy, and let $ϵ > 0$ . Then $\frac{ϵ}{2} > 0$ so using the definition then:

| x_{n + 1} - x_{n} | < \frac{ϵ}{2}, | y_{n + 1} - y_{n} | < \frac{ϵ}{2}

then:

\begin{aligned} | (x_{n + 1} + y_{n + 1}) - (x_{n} + y_{n}) | & \leq | x_{n + 1} - x_{n} | + | y_{n + 1} - y_{n} | \\ < \frac{ϵ}{2} + \frac{ϵ}{2} \\ = ϵ \end{aligned}

Showing $(x_{n} + y_{n})$ is pseudo-Cauchy.
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