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Prove directly that the following sequence diverges to negative infinity

$a_n = -n^2$

I understand that the sequence will diverge to negative infinity. I know that I must somehow integrate an $n$ and $M$ in some sort of inequality. In class we have been doing "scratch works" and then writing the proof starting from the bottom of our scratch work and moving up. Please help as we have done no examples with negative in class.

If the sequence diverges to negative infinity, this implies that for every $M$ < 0 , there exists an N such that $n$ > $N$; this implies that $a_n < M$.

Not sure how to proceed now.

Antonio Vargas
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icedtea
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    For any large, negative value of $M$, can you find $n$ such that $a_n < M$? This would mean $a_n$ marches off toward negative infinity (it is eventually smaller than any negative number). – Austin Mohr Feb 13 '14 at 03:17
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    You have several tags, so it's unclear what level of mathematical sophistication you're looking for. Does the definition of divergence you're using involve epsilons ($\varepsilon$) and $N \in \mathbb{N}$ and such? – Ben Grossmann Feb 13 '14 at 03:17
  • yes we have been using epsilons! This is for my real analysis class and we just started sequences and convergence – icedtea Feb 13 '14 at 03:20
  • You want to show that for any number $-B$, there is an $N$ such that if $n\gt N$, then $-n^2\lt -B$. This is equivalent to saying at $n^2\gt B$, and now we are at positives where some of us (including me) are more comfortable. – André Nicolas Feb 13 '14 at 03:38

1 Answers1

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What we need here is the definition of divergence to $-\infty:$

We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$

So, our proof is going to start as follows:

Fix $M>0$

Sometimes you see "Let $M>0$ be given" but the idea is the same. The game you have to play in this proof is as follows: if I give you an $M$, you have to find an $n_0$ that makes $a_n < -M$ whenever $n>n_0$. If you can do this for any $M$, then you've proven that $a_n$ diverges to $-\infty$.

So here's the next step:

Let $n_0$ be an integer satisfying $n_0 > \sqrt{M}$

So this is where you respond to the challenge by choosing your $n_0$. Of course, given any real number (and $\sqrt{M}$ is an example of such), we can find some integer bigger than that number. This is sometimes referred to as the "Archimedian property" of the real numbers.

All right, now we need to show that this $n_0$ "works". Indeed, we have

For $n>n_0$, we note $$ a_n = -n^2<-n_0^2 < -(\sqrt{M})^2 = -M $$

Great, so now we've won the game! We started with some $M$ (specifying only that it was bigger than zero), and in response we gave an $n_0$ for which $a_n < -M$ whenever $n>n_0$. Since this works for any $M$ (no matter how big), we've shown that $a_n$ diverges to $-\infty$. So, let's finish off:

Thus, we conclude that $\lim_{n\to \infty} a_n = -\infty$

Ben Grossmann
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