Determine whether sequence is non-decreasing or increasing I have a sequence, where $n \in \mathbb N$ $$1-{1\over 3^n}$$
I'm supposed to determine whether the sequence is non-decreasing.
Knowing that for a non-decreasing sequence
$$a_n \le a_{n+1}$$
I'm not sure if my solution is correct
$$1-{1\over 3^n} \le 1-{1\over 3^{n+1}}$$
$$1-{1\over 3^n} \le 1-{1\over 3^{n}\cdot3}$$
$$-{1\over 3^n} \le -{1\over 3^{n}\cdot3}$$
$$-3^n\cdot3 \le -3^n$$
$$3 \ge 1$$
Does this prove the sequence is non-decreasing? If not, how can it be determined?
EDIT:
If I try to prove the sequence is strictly increasing ($a_n \lt a_{n+1}$)
$$1-{1\over 3^n} \lt 1-{1\over 3^{n+1}}$$
$$1-{1\over 3^n} \lt 1-{1\over 3^{n}\cdot3}$$
$$-{1\over 3^n} \lt -{1\over 3^{n}\cdot3}$$
$$-3^n\cdot3 \lt -3^n$$
$$3 \gt 1$$
I am a little confused, since this also proves the sequence is strictly increasing. What am I doing wrong?
 A: What you may also attempt could is called Mathematical Induction. Proof by induction requires 2 things:


*

*Proving that the property $a_n\leq a_{n+1}$ holds for the base case (in this case, $n=1$).

*Proving that the property $a_n\leq a_{n+1}$ holds for any relevant $n$ (in this case, $n\in\mathbb N$).


$a_1 = 1-\frac{1}{3^1} = \frac{2}{3} = \frac{6}{9}$
$a_2 = 1-\frac{1}{3^2} = \frac{8}{9}$
Thus we conclude that $a_1\leq a_2$. 
Adding this to your proof that $a_n\leq a_{n+1}$ for any $n\in\mathbb N$, we've proven the the sequence is non-decreasing.
A: Since “adding $1$” is an increasing operation, you can simply look at the sequence $(-1/3^n)$. This is increasing if and only if the sequence $(1/3^n)$ is decreasing, which is equivalent to $(3^n)$ being increasing, which is true.
Your idea is good as well: in the display below, each line is equivalent to the previous one:
\begin{gather}
1-\frac{1}{3^n}< 1-\frac{1}{3^{n+1}} \\
-\frac{1}{3^n} < -\frac{1}{3^{n+1}} \\
\frac{1}{3^{n+1}}< \frac{1}{3^n} \\
\frac{1}{3} < 1
\end{gather}
Since the last line is a true statement, also the first line is.
A: A more elegant proof can be achieved with a little bit of calculus;
Let $f(n) = 1 - \frac{1} {3^n}$. Then $f'(n) = \frac{1}{3^n}\log(3)$.
Note that $a = \log(3) = 1.099 > 0$, and that for any $n$, $b = \frac{1}{3^n} > 0$. 
Thus, $f(n) = 1 - \frac{1} {3^n} = ab > 0$, which means that $f(n)$ is monotonically increasing, and so it is non-decreasing.
