If $0<a<$1, show that $\inf{a^n}=0$

Question

Since $0$ is a lower bound $\inf{a^n}\geq 0$. How do I prove that $\inf{a^n}=0$?

Answer 1

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.

Answer 2

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 $$ n := \bigg\lfloor \frac{\log \delta}{\log a} \bigg\rfloor + 1. $$

Answer 3

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.

Answer 4

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 $a < 1$?

Answer 5

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 $L = 0$.