# Showing that $a_n = 1-(\frac{-2}{3})^n$ is not bounded

I have the sequence $a_n = 1-(\frac{-2}{3})^n$, and I need to show if it is bounded.

I was first under the assumption that, if a sequence is monotonically increasing/decreasing and bounded that the sequence converges, it also meant that if a sequence converges, there had to be an upper or lower bound. But looking at the graph of the sequence in Wolfram, it turns out to be neither; it is a midpoint, around which the function oscillates, and it oscillates around $\displaystyle{\lim_{n \to \infty} a_n = 1-(\frac{-2}{3})^n} = 1$

So to show that a sequence converges, show that it monotonically increases or decreases and that has a bound. Alright. But how to show that a sequence, like $a_n = 1-(\frac{-2}{3})^n$, is not bounded.

I'm new to all this, so please take it easy ;-)

• $a_n$ is bounded. Its magnitude is always less than 2 e.g. There are more ways to show sequences converge then the way you mentioned. That's a sufficient condition for one type of sequence. – muaddib Jun 25 '15 at 13:04
• a converging sequence is bounded – user190080 Jun 25 '15 at 13:07
• It is? Then I must be doing something wrong on Wolfram - the graph is showing that it is oscillating farther and farther if traveling left on the x-axis. I typed in the sequence as $a_n = 1-(\frac{-2}{3})^n$ in Wolfram. – Garth Marenghi Jun 25 '15 at 13:08
• Consider $a_{2n}$ and $a_{2n+1}$ – Claude Leibovici Jun 25 '15 at 13:08
• Show what did you typed. May be, a typo ? – Claude Leibovici Jun 25 '15 at 13:09

You are right that monotonic, bounded sequences converge, but other sequences converge too. This is an example of a sequence which is not monotonic, but it does converge. You have indicated this yourself in saying that $$\lim_{n\to\infty}a_n=1.$$ Bounded means there exists $M\in\mathbb R$ such that $|a_n|<M$ for all $n$. Since only non-convergent sequences are unbounded, this sequence is bounded.

To prove that $\{a_n\}$ converges to 1, we have to show that we can draw as small an interval around 1 as we would like, and that after a while, the sequence doesn't leave that interval. So we pick the size of the interval. Call this size $\varepsilon$. That means we need to find a number $N$ such that if $n\geq N$, $|a_n-1|<\varepsilon$. Since the sequence $(-2/3)^n$ gets very small as $n\to\infty$, we can pick $N$ which is large enough so that $|(-2/3)^N|<\varepsilon$. Then if $n\geq N$, $|(-2/3)^n|\leq|(-2/3)^N|<\varepsilon$, so $$|a_n-1|=|1-(-2/3)^n-1|=|(-2/3)^n|\leq|(-2/3)^N|<\varepsilon.$$ Thus, for all $n\geq N$, $a_n$ is close to $1$, and the sequence converges.

To answer your other question, to prove that a sequence is unbounded, we have to prove that for all $M\in\mathbb R$, there exists $n$ such that $a_n>M$. This is usually not too difficult (when, indeed, the sequence is unbounded).

Note that $a_n$ is bounded, it is contained in the interval $[1/3,1+2/3]$ because $\left|\left(\frac{-2}{3}\right)^n\right| \leq \frac{2}{3}$ for every $n$

Hints for showing convergence:

1) Show that $\lim_{n\to \infty}\left(\frac{-2}{3}\right)^n = 0$

2) Use the fact that if $\lim_{n\to \infty} x_n = x$ and $\lim_{n\to \infty} y_n = y$ then $\lim_{n\to \infty} x_n -y_n= x-y$

just for the boundedness:

we assume $a_n$ converges to $a$, so therefore we know that $\forall \epsilon >0$ there exists $n_0(\epsilon)$ st

$$\forall n \ge n_0(\epsilon):|a_n-a|<\epsilon$$ so you have your bound for those $a_{n}$, since $a_n\in[a-\epsilon,a+\epsilon]$, then just take the $\max\{|a_n|, n<n_0(\epsilon)\}$ and you're pretty much done.