Does the following sequence converge, if so find the limit, if not prove that no limit exists: $a_n=(-1)^nn^2$

Question

$$a_n=(-1)^nn^2$$

I know that this sequence does not converge and hence does not have a limit. I have tried proving that this sequence does not have a limit by contradiction.

I assumed that the limit 'a' existed and then performed the reverse triangle inequality. Thus:

$$|(-1)^nn^2-a|<\epsilon$$ $$=|(-1)^nn^2|-|a|<\epsilon$$

From here I am stuck, I have been told that I should be considering two cases for the even and odd I believe, but am not too sure how to go forward and make these contradictions.

Answer 1

This is an alternating series. So use the alternating series test.

$$\lim_\limits{n\to \infty} |a_n|=\lim_\limits{n\to \infty}|(-1)^{n}n^2| =\lim_\limits{n\to \infty} n^2 = \infty \not =0$$

Answer 2

If you want to work directly from the definition of a convergent sequence:

Suppose that the sequence converges to $L$. Consider $n=2k$ and $n=2k+1$. For $k$ sufficiently large, $|a_{2k}-L|< \varepsilon$ and $|a_{2k+1}-L|<\varepsilon$.

Therefore, $|a_{2k}-a_{2k+1}|\leq |a_{2k}-L|+|a_{2k+1}-L|<2\varepsilon$. Now, choose a $k$ so large that this is impossible.

Answer 3

There are different ways to see that it does not converge. One of the simplest ways is to note that your sequence $a_n$ is unbounded: It tends to $\infty$ along all even numbers $2n$, and tends to $-\infty$ along all odd numbers $2n+1$. Since a convergent sequence is bounded, this gives the result.

Answer 4

Hint: Recall that $(-1)^n$ does not converge and product of two convergent sequences is again convergent.