Let $a ∈ (0,1).$ Prove that the infimum of the set $\{a^n : n ∈\Bbb N\}$ is zero.

Question

Let $a ∈ (0,1).$ Prove that the infimum of the set $A:=\{a^n : n ∈\Bbb N\}$ is zero.

Can someone explain me step by step how to solve it?

I understand that the sequence converges $0,$ then I have the intuition, but iIreally don't know how to do it.

Answer 1

The sequence $(a^n)$ is a decreasing positive sequence, so it converges: $a^n \downarrow L\geq 0$.

Since $(a^n)$ is decreasing, $\inf a^n = \lim a^n = L$.

Since $a^n\downarrow L$, then also $a\cdot a^n\downarrow aL$, i.e., $a^{n+1}\downarrow aL$.

But $(a^{n+1})$ is a subsequence of $(a^n)$, so they have the same limit: $L = aL$.

Since $a\neq 0$, this implies $L = 0$.

Answer 2

Following MPW argument, you can do as follows.

Let $a\in(0,1)$, and the secuence $\{a^n\}_{n\in\mathbb{N}}$. Clearly, $0<a\implies 0<a^2\implies 0<a^3\dots$, so by induction $0<a^n\ \ \forall n\in \mathbb{N}$. Moreover, as $a<1$, by induction again, we obtain $a^{n+1}<a^n\,\,\forall n\in\mathbb{N}$. Therefore, we have a decreasing monotone sequence bounded from below, so we know it is convergent.

Now we can apply MPW argument to see that as $\{a^n\}_{n\in\mathbb{N}}$ is indeed convergent to a real number $L$, that number verifies $L=aL$ and as $a\not=1$, this implies $L=0$. So we have deduced that $\underset{n\to\infty}{lim}\ a^n=0$.

Now, back to the original problem. The infimum of a set X is a number $u$ that is a lower bound for $X$ and such that if $\varepsilon>0$, then $u+\varepsilon$ is not a lower bound for $X$. From what we deduced on the first paragraph, $0$ is a lower bound for $A$. Now let $\varepsilon>0$, we consider the number $0+\varepsilon=\varepsilon$, and see that this is not a lower bound for $A$. Indeed, as $\underset{n\to\infty}{lim}\ a^n=0$, for all $\varepsilon>0$ there exists sufficiently large $n_0\in\mathbb{N}$ such that $|a^n-0|<\varepsilon\,\,\forall n\geq n_0$. As $a^n>0\,\,\forall n$, we deduce that there exist a natural number, namely $n_0$ such that $a^{n_0}<\varepsilon$, and therefore, $\varepsilon$ is not a lower bound for $A$. This finishes the argument, and $inf(A)=0$.

But notice how this argument provides an even stronger result (because in your case A was countable set representing precisely the sequence $\{a^n\}_{n\in\mathbb{N}}$), as it can be used to see that if we consider the set $B=\{a^n\ ||\ a\in(0,1),\, n\in\mathbb{N}\}$, which is not countable at all and very far away from being a sequence, then the argument done allow us to inmediately deduce $inf(B)=0$.