prove that $\inf \{ a^n:n\in \mathbb{N}^* \}=0, 0<a<1$

Question

$$\inf \{a^n:n\in\mathbb{N}^*\}=0,\ \ 0<a<1$$

What i have managed so far:

since $n>0$ then $a^n>0$ Let $ε>0$

after that i m not really sure what to do in order to prove $\inf \{a^n:n\in\mathbb{N}^*\}=0$...help with explanation would be much appreciated.

Answer 1

Since $0<a<1$, $\frac1a>1$. Therefore$$\frac1{a^n}=\left(\frac1a\right)^n=\left(1+\left(\frac1a-1\right)\right)^n\geqslant 1+n\left(\frac1a-1\right),$$by Bernoulli's inequality. In particular,$$\frac1{a^n}>n\left(\frac1a-1\right).$$By the Archimedean property,$$n>\frac1{\left(\frac1a-1\right)\varepsilon}$$if $n$ is large enough. So$$\frac1{a^n}>n\left(\frac1a-1\right)>\frac1\varepsilon$$which is equivalent to $a^n<\varepsilon$.

Answer 2

Sketch of proof (I'll provide more details upon request):

Show that $(a^n)$ is a monotonically decreasing sequence, which is bounded below (by zero). Therefore it converges to its infimum, which I will call $L$.

Then, writing $x_n = a^n$, observe that $$L = \lim_{n \to \infty}x_n = \lim_{n \to \infty}x_{n+1} = \lim_{n \to \infty}a^{n+1} = a\lim_{n \to \infty}a^n = aL$$ The second equality holds because if a sequence converges, then any subsequence converges to the same limit.

What can you conclude from $L = aL$?

Answer 3

Since $a^n>a^{n+1}$ for all $n\in N^*$, finding the infimum is equivalent to finding the limit of the sequence $x_n=a^n$. This sequence can be written recursively: $$x_1=a, \\x_{n+1}=ax_n.$$ Since $a<1$, the sequence is monotonously decreasing. Since $a^n>0$, as you noticed, the sequence is also bounded, hence convergent. Let $x\in \Bbb R$ be the limit. Then $x$ must satisfy the recursion: $$x=ax.$$ But, since $a\neq0$, this only holds for $x=0$.

Answer 4

No mention of what you know... In particular for Bungo's solution you have to know that multiplication is continuous. That's why José's solution is great.

Assuming we know whatever we want, use logarithm and exponential:
$0<a<1$ so $\log(a) < 0$, then by Archimedian property $n\log a \to -\infty$ as $n \to \infty$, so exponentiate to conclude $a_n \to 0$.

Answer 5

Since

$0 < a < 1, \tag 1$

we have

$a^n > 0, \; \forall n \in \Bbb N^\ast; \tag 2$

also,

$a^{n + 1} < a^n, \; \forall n \in \Bbb N^\ast. \tag 3$

We see from (2), (3) that the sequence $a^n$ is monotonically decreasing and bounded below by $0$; it follows that there exists $\epsilon$, $0 \le \epsilon < 1$, with

$\epsilon = \inf \{ a^n, \; n \in \Bbb N ^\ast \} = \lim_{n \to \infty} a^n; \tag 4$

now if

$\epsilon > 0, \tag 5$

then

$0 < a \epsilon < \epsilon. \tag 6$

Now consider the continuous function $f: \Bbb R \to \Bbb R$ given by

$f(x) = ax; \tag 7$

the continuity of $f(x)$ implies

$\lim_{n \to \infty} f(a^n) = f( \lim_{n \to \infty} a^n) = f(\epsilon) = a\epsilon < \epsilon; \tag 8$

however,

$f(a^n) = aa^n = a^{n + 1}, \tag 9$

whence

$\lim_{n \to \infty} f(a^n) = \lim_{n \to \infty} a^{n + 1} = \epsilon; \tag{10}$

we thus see that, unless $\epsilon = 0$, (8) is inherently self-contradictory; thus we must have $\epsilon = 0$ and

$\inf \{ a^n, \; n \in \Bbb N^\ast \} = 0, \tag{11}$

as was to be shown.