Prove $\{n^2\}_{n=1}^{\infty}$ is not convergent

Question

Definition of convergent sequence:

There exists an $x$ such that for all $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, $d(x_n, x) < \epsilon$.

So the negation is that there does not exist an $x$ such that for all $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, $d(x_n, x) < \epsilon$.

The negation doesn't really help me here, since we weren't given a specific limit point of the sequence to disprove, and we have to prove this for all $x \in \mathbb{R}$. I have never done this before since we were always given some limit point along with the sequence. What is the way to prove this for all $x$?

Answer 1

Suppose to the contrary that the sequence converges to a limit $L$. Then, for all $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that

$$ |n^2-L|<\varepsilon $$ for every $n\geq N$. Fix $\varepsilon<1/2$. For any $n\geq N$, $$ 1>2\varepsilon > |n^2-L|+|L-(n+1)^2| \geq |n^2-L+L-(n+1)^2| = |n-(n+1)^2|>1, $$ which is a contradiction. Thus, the sequence does not converge.

EDIT: Typically whenever you want to prove the divergence of a sequence (this is an easier case since it is not bounded, as some people point out) you can assume that it converges, so that you have freedom to choose a suitable $\varepsilon$ that will lead to contradiction. For example, in this case $\varepsilon<1/2$ was OK, and for the sake of completeness, if you consider the sequence $(-1)^n$ (which is bounded) then some $\varepsilon\leq 1$ would make the job. You can easily guess these $\varepsilon$ by looking at $|a_n-a_{n+1}|$.

Answer 2

The definition of convergence is

"There exists a $x\in\mathbb{R}$, such that for all $\epsilon>0$, there exists an $N$ such that for all $n\ge N$, $d(x_n,x)<\epsilon$"

The negation is

"There is no $x\in\mathbb{R}$, such that for all $\epsilon>0$, there exists an $N$ such that for all $n\ge N$, $d(x_n,x)<\epsilon$"

Since $n^2$ is monotone crescent and is not bounded (there is no maximum), for all $x\in\mathbb{R}$ there is a $n$ large enough such that $d(x_n,x)=|n^2-x|>\epsilon$

Answer 3

Since $\mathbb{R}$ is a complete metric space, every convergent sequence is bounded.

$\{n^2\}_{n\geq 1}$ is not a bounded sequence, hence it cannot be convergent.