Give an example of each or state that the request is impossible. For a

Charles Kingsley 2021-12-17 Answered
Give an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case.
a) A sequence with an infinite number of ones that does not converge to one.
b) A sequence with an infinite number of ones that converges to a limit not equal to one.
c) A divergent sequence such that for every \(\displaystyle{n}\in{N}\) it is possible to find nn consecutive ones somewhere in the sequence.

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Expert Answer

Kayla Kline
Answered 2021-12-18 Author has 2853 answers

Step 1
Take sequence \(\displaystyle{x}_{{{n}}}={\left(-{1}\right)}^{{{n}}}\) where 1 occurs infinitely many times but does not converge to 1.
This is because if we choose \(\displaystyle\epsilon\le{2}\), given any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), where exists \(\displaystyle{n}={2}{N}+{1}\) such that
\(\displaystyle{\left|{\left({1}\right)}^{{{n}}}-{1}\right|}={2}\geq\epsilon\)
Step 2
b) Suppose \(\displaystyle{x}_{{{n}}}\) is a convergent sequence with infinitely many 1 but converges to \(\displaystyle{x}\ne{1}\). Let \(\displaystyle\epsilon={\left|{1}-{x}\right|},\) then for any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), there exist \(\displaystyle{n}\geq{N}\) such that \(\displaystyle{x}_{{{n}}}={1}\). So
\(\displaystyle{\left|{x}_{{{n}}}-{x}\right|}={\left|{1}-{x}\right|}\geq\epsilon\)
So this is not possible
Step 3
c) Consider a sequence \(\displaystyle{\left({x}_{{{n}}}\right)}\) given by
\(x_{n}=\begin{cases}r & if\ n=\frac{r(r+1)}{2}-1\ \text{for some}\ r\in\mathbb{N}\\ 1 & \text{otherwise}\end{cases}\)
(or put any thing in place of r other than 1)
There n consecutive 1's occur in the sequence. Some of the terms in sequence are
\(\displaystyle{\left({x}_{{{n}}}\right)}_{{{n}\in{\mathbb{{{N}}}}}}={\left({1},\ {2},\ {1},\ {1},\ {5},\ {1},\ {1},\ {1},\ {9},\ {1},\ {1},\ {1},\ {1},\ {14},\ \cdots\right)}\)
This is constructed this way
\(\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots\end{matrix}\)

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scoollato7o
Answered 2021-12-19 Author has 158 answers

Step 1
a) Consider the sequence \(\displaystyle{x}_{{{n}}}={\left(-{1}\right)}^{{{n}}}\)
Here the occurrence of 1 is infinite times and it does not converge to 1.
That is, on choosing \(\displaystyle\epsilon\le{2}\), given any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), there exist \(\displaystyle{n}={2}{N}+{1}\) such that \(\displaystyle{\left|{\left({1}\right)}^{{{n}}}-{1}\right|}={2}\geq\epsilon\)
Step 2
b) Consider a convergent sequence \(\displaystyle{\left({x}_{{{n}}}\right)}\) with infinitely many 1 but it converges to \(\displaystyle{x}\ne{1}\).
Let \(\displaystyle\epsilon={\left|{1}-{x}\right|}\) then for any \(\displaystyle{N}\in{\mathbb{{{N}}}}\), there exist \(\displaystyle{n}\geq{N}\) such that \(\displaystyle{x}_{{{n}}}={1}\). So,
\(\displaystyle{\left|{x}_{{{n}}}-{1}\right|}={\left|{1}-{x}\right|}\geq\epsilon\)
Thus, this is not possible.
Step 3
c) Consider a sequence \(\displaystyle{\left({x}_{{{n}}}\right)}\) given by
\[x_{n}=\begin{cases}r & if\ n=\frac{r(r+1)}{2}-1\ \text{for some} r\in\mathbb{N}\\ 1 & \text{otherwise}\end{cases}\]
There exist n consecutive 1's occur in the sequence. Some of the terms in the sequence will be
\[(x_{n})_{n\in\mathbb{N}}=(1,\ 2,\ 1,\ 1,\ 5,\ 1,\ 1,\ 1,\ 9,\ 1,\ 1,\ 1,\ 1,\ 14,\ \cdots)\]
Which gives a contradiction.
That is,
\(\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots \end{matrix}\)
Thus, it is impossible.

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content_user
Answered 2021-12-26 Author has 9769 answers

Step 1
a) Consider the sequence
\(\{a_{n}\}=\{(-1)^{n}\}=\{-1,\ 1,\ -1,\ 1,\ \cdots\}\)
\(a_{n}\) has an infinite number of ones. Indeed, if n is even, \((-1)^{n}=1\), and there are infinite even numbers, hence infinite ones. However, \(\{a_{n}\}\) does not converge to one, because \(\{a_{n}\}\) is divergent (it oscillates between 1 and -1).
b) This is impossible. We will use the definition of limit to see why.
Let \(\epsilon>0\). Suppose that \(\{a_{n}\}\) has an infinite number of ones, and suppose that \(\{a_{n}\}\) converges to L. Then, there exists some M>0 such that if \(m\geq M\) then \(|a_{n}-L|<\epsilon\). Now, there exist infinite natural numbers \(n\geq M\) such that \(a_{n}=1\) by hypotheses. Then \(|1-L|<\epsilon\)
Hence L and 1 get arbitrarily close, therefore L=1
c) It is possible to construct such a sequence:
\(\{a_{n}\}=\{1,\ 0,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 1,\ 0,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 0,\ \cdots\}\)
even if there is not a simple explicit formula. The sequence is divergent because it has infinite zeroes and infinite ones (thus L should be either 0 or 1, but it can't be both).

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