If $0Since $0$ is a lower bound $\inf{a^n}\geq 0$. How do I prove that $\inf{a^n}=0$?
 A: Prove: if $S$ is a non-empty set of real numbers with a lower bound, so that $\inf S$ is defined, and if $aS$ denotes the set $\{ax : x \in S\}$, then $\inf(aS)$ is also defined, and because $a > 0$, $\inf(aS) = a(\inf S)$. Observe that if $S = \{a, a^2, a^3, \ldots\}$, then $aS \subset S$; what now follows, in view of the fact that $a < 1$?
A: Hint: Suppose $\inf a^n=b>0$.  Then for any $\epsilon>0$, there exists $n$ such that $a^n<b+\epsilon$.  Try to choose $\epsilon$ so that this would imply $a^{n+1}<b$ to get a contradiction.
A: Since
$0 < a < 1$,
$a = \frac1{1+b}
$
where
$b = \frac1{a}-1 > 0$.
Then,
by Bernoulli's inequality,
$(1+b)^n \ge
1+bn
> bn
$
so
$a^n
=\frac1{(1+b)^n}
< \frac1{bn}
= \frac1{n(1/a-1)}
\to 0
$.
Note:
This is not original.
I saw this in
"What is Mathematics?"
by Courant and Robbins,
a book I highly recommend.
A: We are to prove that for every $\delta > 0$ there is some $n \geq 1$ such that $0 \leq a^{n} < \delta$. The left inequality is trivial, for by assumption $0 < a < 1$. It remains to prove the right inequality. Let $\delta > 0$. If $\delta \geq 1$, then $a^{n} < \delta$ for all $n \geq 1$; suppose $\delta < 1$. Then $a^{n} < \delta$ iff
$$
n|\log a| > |\log \delta|,
$$
iff
$$
n > \frac{\log \delta}{\log a}.
$$
So there is some $n \geq 1$ such that $a^{n} < \delta$, say 
$$
n := \bigg\lfloor \frac{\log \delta}{\log a} \bigg\rfloor + 1.
$$
A: First note that $a > 0$, so for any positive integer $n$ we have $a^n > 0$. We also have $a < 1$. Multiply both sides of this inequality by $a^n$ to obtain $a^{n+1} < a^n$. This shows that $x_n = a^{n}$ is a decreasing sequence which is bounded below (by zero), hence it is convergent to its infimum, which we will call $L$.
Now note that on one hand,
$$\lim_{n \to \infty}a^{n+1} = a\lim_{n \to \infty}a^n = aL$$
But on the other hand,
$$\lim_{n \to \infty}a^{n+1} = \lim_{n \to \infty}x_{n+1} = \lim_{n \to \infty}x_n = \lim_{n \to \infty}a^n = L$$
where the second equality holds because $\{x_{n+1}\}$ is a subsequence of $\{x_n\}$.
Since the LHS of the above two chains of equalities are the same, the RHS must also be equal, hence $aL = L$, which means that $L(a-1) = 0$. As $a \neq 1$, this forces $L = 0$.
