Prove $\forall n \in\Bbb N$, $0 < a < 1$ $\implies$ $a^n \leq 1$

Question

I'm trying to prove this by induction but I'm running into some trouble.

The base case is $0$, so, $a^0 = 1$, the inequality holds true

Being new to induction, I don't exactly know what to do for the A and proving it in the inequality.

at this point, do I just prove $a^n \le 1$ implies $a^{n + 1}\le 1$?

I don't understand how the $0 < a < 1$ changes things.

Please help!

Answer 1

Yes, you want to show that given $a^n\le 1$, then $a^{n+1}\le 1$. Well since $a>0$, we know that $a^{n+1}\le a$. But $a<1$. So we have $a^{n+1}\le a\le1$ as desired.

This proof would clearly not work if $a$ were not less than $1$.

Answer 2

Lemma: If $0<a<1$ I claim for any $b>0$ that $0<ab<b$

Proof: $b(1-a)$ is a product of positive numbers hence is positive.

Corollary: $a^n<1\implies a^{n+1}<1$ when $0<a<1$

Proof: By induction, the base case being given, then let $b=a^n$ in the lemma.

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Alternatively, a direct proof:

Write

$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+\ldots +a+1)$$

Then as the latter element of the product is positive--since $a>0$ (note this means $a^n-1>0$ which is equivalent to $a^n>1$)--the whole thing is positive if $a>1$ and negative if $0<a<1$.

Answer 3

$0 < a < 1 \ \Rightarrow \ 1 = a + x$ with $x > 0$. Thus, $$ 1 = 1^n = (a + x)^n = a^n + \sum_{k=0}^{n-1}{n \choose k}a^kx^{n-k} \geq a^n \quad \Rightarrow \quad a^n \leq 1 $$

Answer 4

Proof: by mathematical induction Choose any positive real number a which is smaller than 1, then

0

Because of (1), we conclude that 0

Good luck