# Proving that if $a \le -1$ then $\underset{n \to \infty}{\lim} a^n$ does not exist

I am trying to show that if $a \le -1$ then $\underset{n\to \infty}{\lim} a_n$ where $a_n = a^n$ does not exist.

I have already ruled out the cases of negative and positive infinity.

Now supposing that such a limit does exist as a real number, let $\underset{n\to \infty}{\lim} a_n = J$ such that $|a_n - J| \le \epsilon$ for any $\epsilon > 0$ for some $n>N$.

How can I proceed to find a contradiction?

• What do you mean by $|a|\leq -1$? The absolute value is by definition positive definite. – Ethan MacBrough Jun 19 '16 at 2:18
• @EthanZed mistake edited. – IntegrateThis Jun 19 '16 at 2:20
• Try proving that if $b_n$ is convergent, then $|b_n|$ is convergent (then use the contrapositive). – Cameron Williams Jun 19 '16 at 2:23
• In addition to Cameron's tip, for the special case of $a=-1$ it's easiest to show the sequence is not Cauchy. – Ethan MacBrough Jun 19 '16 at 2:27
• @EthanZed Good call. I missed that case. – Cameron Williams Jun 19 '16 at 2:27

Suppose $J$ is the limit of $a^n$. Then for any $\epsilon$, we have $|a^n-J|< \epsilon$ for all $n>N$ for some $N$. Then given $\epsilon$, choose $N$ such that $|a^n-J|<\epsilon/|a|$ for $n>N$. Then $|a|\cdot|a^n-J|=|a^{n+1}-aJ|<\epsilon$ for $n>N$. Hence $\lim_{n\to\infty} a^n =J = aJ$. Then $J=aJ$, so either $J=0$, or $J=1$. Of course we know since $a\le -1$, $a1\ne 1$, so we must have $J=0$. But $|a^n| = |a|^n$ is monotonically increasing and always greater than one. Since $|\cdot|$ is continuous, if $\lim_{n\to\infty}a^n=0$, then we would have $\lim_{n\to\infty}|a^n|=|0|=0$. Contradiction.

Alternatively, and perhaps more simply if you are comfortable with the definition of a Cauchy sequence, $|a^{n+1}-a^n|=|a|\cdot|a^n-a^{n-1}|\ge |a^n-a^{n-1}|$. Hence the sequence cannot be Cauchy and cannot converge. (You do have to check that some difference is nonzero, which is why this argument works for $a=-1$, but not $a=1$.)

Here's a useful lemma:

Lemma. If $(b_n)$ is a convergent sequence of real numbers, then $(|b_n|)$ is a convergent sequence of real numbers.

Proof: Let $\varepsilon > 0$, then there exists an $N$ such that for all $n,m>N$, $|b_n-b_m|<\varepsilon$. By the reverse triangle inequality, we have that $||b_n|-|b_m|| < |b_n-b_m| < \varepsilon$, i.e. $(|b_n|)$ is convergent.

In your case, if $a < -1$, let $b_n = a^n$. Then $|b_n| = |a|^n$. Is this convergent? What does that then say about the convergence of $(b_n)$? The case of $a=-1$ is straightforward.

If $a = 1$ then the sequence $a^{n} = (-1)^{n}$ is clearly oscillating between $-1$ and $1$ and hence $\lim_{n \to \infty}a^{n}$ does not exist. Now let $a < -1$ and but $b = -a > 1$ and then we have $a^{n} = (-1)^{n}b^{n}$. Consider the sequence $b^{n}$ which is strictly increasing because $b > 1$. Hence either $b^{n} \to L > 1$ or $b^{n} \to \infty$ as $n \to \infty$. If $b^{n} \to L$ then $b^{n + 1} = b\cdot b^{n}$ implies that $L = b \cdot L$ and this is not possible because both $L, b$ are greater than $1$ and hence $bL > L$. It now follows that $b^{n} \to \infty$. It then follows then $a^{n} = (-1)^{n}b^{n}$ oscillates infinitely and hence $\lim_{n \to \infty}a^{n}$ does not exist.