 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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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
$$\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots\end{matrix}$$

Not exactly what you’re looking for? scoollato7o

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
$(x_{n})_{n\in\mathbb{N}}=(1,\ 2,\ 1,\ 1,\ 5,\ 1,\ 1,\ 1,\ 9,\ 1,\ 1,\ 1,\ 1,\ 14,\ \cdots)$
That is,
$$\begin{matrix} 1 & 2 \\ 1 & 1 & 5 \\ 1 & 1 & 1 & 9 \\ \cdots \end{matrix}$$
Thus, it is impossible. content_user

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).