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How to prove that the sequences $(n)_{n\in\mathbb{N}}$ and $(\frac1n)_{n\in\mathbb{N}^*}$ are not constant?

Using the defintion, a sequence $(u_n)$ is eventualy constant if and only if

$$\exists n_0\in\mathbb{N}, a\in\mathbb{R}\text{ such that } \forall n\in \mathbb{N}, n\geq n_0\Rightarrow u_n=a.$$

Thank you.

3 Answers 3

1

Assume that the sequence $(n)$ is eventually constant.

So there are an $n_0$ and an $a$ such that $\forall n\ge n_0:n=a$. Then taking $n_0$ and $n_0+1$, we have $n_0=a\land n_0+1=a$, which is contradictory.

Similarly with $\dfrac1{n_0}=a\land\dfrac1{n_0+1}=a$ (and any sequence such that $u_{n_0}\ne u_{n_0+1}$).

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You have for the first sequence

$$1\ne 2$$

so it is not constant.

You have for the second sequence

$$\frac 11\ne \frac 12$$

so it is also not constant.

Edit

Since you mean eventually constant instead of constant, for the first one you can notice that

$$\forall n,\quad n

so it won't be eventually constant.

And for the second one you can notice that

$$\forall n,\quad \frac 1n>\frac 1{n+1}$$

so it also won't be eventually constant.

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    i know this, i mean how to use the defintion , a sequence $u_n$ is constant iff $\exists n_0\in\mathbb{N}, \forall n\in \mathbb{N}, n\geq n_0\Rightarrow u_n=a$ – 2017-01-26
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    @Vrouvrou This is not the same question. You mean *eventualy* constant. – 2017-01-26
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    yes eventualy constant – 2017-01-26
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    what is $n_0$ in the two case, i know what you are writing but i want to apply the definition, and find that is not true – 2017-01-26
1

The statement you want to disprove is the statement

$\exists a\in\mathbb R,\ \exists n_0\in\mathbb N,\ \forall n\in\mathbb N: n\ge n_0\implies u_n=a$

The negation of that statement is

$\forall a\in\mathbb R,\ \forall n_0\in\mathbb N,\ \exists n\in \mathbb N: u_n\neq a\land n\ge n_0$

This statement is easy to proves, since you can take any $a$ and any $n_0$, and then take $n=n_0$ or (if $u_{n_0}=a$) $n=n_{0}+1$