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

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 $x$?

• Can you correct your title? something is wrong with \limits – Jimmy R. Mar 31 '15 at 12:12
• Use \lim instead of \limits – Fermat Mar 31 '15 at 12:16
• Do you mean $(n^2)_{n=1}^\infty$ in the title? – Hagen von Eitzen Mar 31 '15 at 12:20
• Actually, your first definition is not "The sequence $(x_n)$ is convergent", but rather it is "The sequence $(x_n)$ converges to $x$". The definition of convergent should have another quantor for $x$, i.e. "There eixts $x$ such that the sequence converges to $x$". – Hagen von Eitzen Mar 31 '15 at 12:23
• Considering @HagenvonEitzen comment, you should add a quantor for $x$ and therefore you must prove that for all $x\in \mathbb{R}$. the sequence doesn't converge to $x$, which is a little bit different from what you're trying. What you're proving is that your sequence is not a Cauchy sequence so it doesn't converge, this approach is easier and it doesn't involve any particular limit $x$. – Daniel Mar 31 '15 at 12:27

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 $|a_n-a_{n+1}|$.

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.

• How do you prove it's not bounded if you are not given a particular bound to disprove? – Carsten S Mar 31 '15 at 12:54
• @CarstenSchultz: just take any possible upper bound and disprove it: given $a_n=n^2$ and a real number $r\in\mathbb{R}^+$, for every $n>\sqrt{r}$ we have $a_n>r$, hence there is no possible working upper bound. – Jack D'Aurizio Mar 31 '15 at 13:23
• Thanks for the answer, and my apologies for not being clearer in my comment, but that seems to be very similar to the thing that the OP has problems doing for convergence instead of boundedness. – Carsten S Mar 31 '15 at 13:26

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 $d(x_n,x)=|n^2-x|>\epsilon$