$(a_{n})_{n=1}^{\infty}$ is a convergent sequence of integers. Prove the existence of $N \in \mathbb{N}$ such that $a_i = a_j$ for all $i, j > N$.

Question

Assume $(a_{n})_{n=1}^{\infty}$ is a convergent sequence of integers. Prove the existence of $N\in\mathbb{N}$ such that $a_i = a_j$ for all $i, j > N.$

I am completely lost on what this question is asking, do I use the definition of a limit to show it converges and doesn't leave epsilon? I have searched and can't find this precise question.

score 2 · Answer 1 · answered Mar 13 '19 at 10:23

2

Let $a_n \to a$. Then there exist $N$ such that $|a_n-a_m| \leq |a_n-a|+|a_m-a| <\frac 1 2+\frac 1 2$ for all $n,m \geq N$. But $a_m$ and $a_m$ are integers, so $|a_n-a_m| <1$ implies $a_n=a_m$ for all $n,m \geq N$.

answered Mar 13 '19 at 10:23

Kavi Rama Murthy

311,013

score 1 · Answer 2 · answered Mar 13 '19 at 10:25

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If $(a_n)$ is convergent, then $(a_n)$ is Cauchy. Hence there is $N \in \mathbb N$ such that

$|a_n-a_m|<1$ for $m,n > N$. This gives $a_n=a_m$ for $m,n > N$, since all $a_n \in \mathbb Z.$

answered Mar 13 '19 at 10:25

Fred

77,394

score 0 · Answer 3 · answered Mar 13 '19 at 10:45

0

Option.

Since $a_n \in \mathbb{Z^+}$, $n=1,2,3,.....,$ is convergent in $\mathbb{R}$ it is Cauchy.

Let $\epsilon <1$ given.

There is a $N$ s.t. for $n \ge m\ge N$

$|a_n-a_m| <1$ , since $a_n,a_m$ are integers we have

$a_n=a_m$.

answered Mar 13 '19 at 10:45

Peter Szilas

20,344
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is an and am simply referring to the ai and aj in the question? – john smith Mar 13 '19 at 10:48
@johnsmith Yes, exactly. – Pedro Mar 13 '19 at 10:55
John.Yes! My comment disappeared.Pedro answered already.n,m are dummies.You can rewrite the whole thing with i j, instead.Ok? – Peter Szilas Mar 13 '19 at 10:58

$(a_{n})_{n=1}^{\infty}$ is a convergent sequence of integers. Prove the existence of $N \in \mathbb{N}$ such that $a_i = a_j$ for all $i, j > N$.

3 Answers3