2
$\begingroup$

Prove $\displaystyle\lim_{n\to\infty} n=\infty$ using the limit definition for a converging sequence.

What I did: Suppose by contra position that $n$ tends to a finite real limit $L$, so from the definition of a converging sequence we have: $|n-L|<\epsilon\Rightarrow n<\epsilon+L$ so if we'll choose $N$ to be $N=\epsilon+L$ then because by definition the above is true for all $n>N$ we'll have: $\epsilon+L+1<\epsilon+L$ contradiction.

But apparently it's wrong and I was told I can't simply choose $N$, but why not? The definition states such $N$ exists...

Note: we use a definition with a real $N$.

$\endgroup$
1
  • 3
    $\begingroup$ This argument only shows the limit is not converging to a real number. The sequence $(1,-1,1,-1,...)$ also does not converge to any real number, and $\lim\limits_{k\to\infty} (-1)^k\neq\infty$. $\endgroup$
    – Ian Mateus
    Dec 14, 2014 at 13:34

4 Answers 4

2
$\begingroup$

I suppose you mean contradiction, not contraposition.

As Ian Mateus points out in the comments, assuming it converges to a finite number is not the negation of diverging to infinity.

You can prove it directly. By definition, $$\lim_{n\to\infty}n=\infty \iff \forall M\in ]0,+\infty[\exists N\in \mathbb N\forall n\in \mathbb N(n\ge N\implies n>M).$$ So take $M$ as above, let $N$ be a natural number greater than $M$ and it follows easily.

$\endgroup$
7
  • $\begingroup$ Wait so the answer by user38584 is wrong? $\endgroup$
    – shinzou
    Dec 14, 2014 at 13:39
  • $\begingroup$ @kuhaku I commit to this: none of the other answers at the moment proves that $\lim \limits_{n\to \infty}(n)=\infty$. $\endgroup$
    – Git Gud
    Dec 14, 2014 at 13:41
  • $\begingroup$ No kidding, they actually proved it in our lecture almost exactly as user38584 did... $\endgroup$
    – shinzou
    Dec 14, 2014 at 13:44
  • 1
    $\begingroup$ Git Gud is absolutely right, proving the result is easy, but certainly not acieved by showing there is no finite limit. $\endgroup$ Dec 14, 2014 at 13:49
  • 1
    $\begingroup$ @kuhaku Using just the definition of converging (to a real number) sequence, it's not possible. It misses cases like $\left(\sin \left(\frac{k\pi}2\right)\right)_{k\in \mathbb N}$. $\endgroup$
    – Git Gud
    Dec 14, 2014 at 14:04
2
$\begingroup$

You can not simply choose an $N$ because that is not what the definition states. It says for any $\varepsilon>0$, there exists an $N\in\mathbb{N}$ such that $|x_n-L|<\varepsilon$ for all $n\ge N$. That means you have to choose an $\varepsilon$ and then find an $N$.

$\endgroup$
2
  • $\begingroup$ Yes, I agree with this. I was a little confused with what the question was asking. $\endgroup$
    – Math1000
    Dec 14, 2014 at 13:27
  • $\begingroup$ Aha, I see. Well thanks. $\endgroup$
    – shinzou
    Dec 14, 2014 at 13:27
2
$\begingroup$

The basic question is how do you know that there exists at least an $n\in \mathbb{N}$ such that $n>N$? For this you need to prove the unboundedness of $\mathbb{N}$ which is what you are asked to prove. So basically your argument is circular and hence wrong.

$\endgroup$
2
  • $\begingroup$ Can't I assume that $n>N$ is true from the limit definition? $\endgroup$
    – shinzou
    Dec 14, 2014 at 13:32
  • $\begingroup$ No. You can't. For this you have to prove that at least such an $n$ exists which would satisfy the inequality. $\endgroup$
    – user170039
    Dec 14, 2014 at 13:34
1
$\begingroup$

Suppose the sequence $x_n = n$, $n=1,2,3,\ldots$ has a limit $L$. Then for any $\varepsilon > 0$, there exists $N$ such that $n\geqslant N$ implies $|x_n-L|<\varepsilon$. So take $\varepsilon = 1$ and choose such an $N$. Let $m = 2 + \max\{N, \lceil L \rceil\}$ (where $\lceil\cdot\rceil$ denotes the ceiling function). Then $m\geqslant N$ but $m-L\geqslant2>1=\varepsilon$, a contradiction.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.