Proving limit of $a^n\to0$ for $|a|<1$ without use of logarithms

Question

Prove that $a^n\to0$ as $n\to∞$ for $|a|<1$ without use of logarithms by using properties of the sequence $u_n=|a|^n.$

I've noticed that I should use the subsequence $u_{2n}$, and the fact that $u_{2n}=u_n^2$. However, I don't know where to go from here. I'm not familiar with these types of proofs so a hint/solution would be greatly appreciated.

Thank you!

Answer 1

Just for simplicity consider the case $0<a<1.$ Now consider the sequence $u_n=a^n.$ Show that

1. $u_n$ is decreasing;
2. $u_n$ is bounded from below.

Thus we have that $u_n$ is convergent. Let's write $L=\lim_n u_n.$ Since $u_{2n}=u_n^2$ we have that

$$\lim_n u_{2n}=L=L^2=\lim_n u_n^2.$$ So $L=L^2$ from where $L=0$ or $L=1.$ The case $L= 1$ is not possible since $u_n<1$ and it is decreasing.

Answer 2

Assume $0\le a<1$ and set $b=\frac1a >1$ then by Bernoulli inequality

$$b^n=(1-(b-1))^n\ge 1-n(b-1) \to \infty$$

and then

$$a^n=\frac1{b^n}\to 0$$

More in general for $|a|<1$ we have that

$$-|a|^n\le a^n\le |a|^n$$

and then by squeeze theorem we conclude that

$$a^n \to 0$$

Answer 3

If $|a|<1$, then $1/|a| >1$, hence there is $x>0$ such that $\frac{1}{|a|}=1+x$. With Bernoulli we get for $ n \in \mathbb N$:

$\frac{1}{|a^n|}=(1+x)^n \ge 1+nx > nx$.

Hence

$|a^n| <\frac{1}{nx}$.

Conclusion: $a^n \to 0$.

Answer 4

Note that $$ |a^{n+1}| = |a||a^n|$$

The sequence $\{|a^n|\}$ is decreasing and bounded below so it converges to $L$

Taking limit of $$ |a^{n+1}| = |a||a^n|$$ we get $|a|L=L$ which implies $L=0$

Answer 5

Why not just (assuming as mfl did that $0<a<1$) use definitions: $\{a^n\}$ is decreasing and bounded below, hence convergent, as in mfl's post, and the limit $L = \inf a^n$ is pretty clearly nonnegative. Assume by way of contradiction that $L>0$. Then there exists $N>0$ such that $n\ge N$ implies $a^n < L(1/a)$. Then $a^{N+1} < L$, contradiction.

Answer 6

By ratio test

$$\frac{|a|^{n+1}}{|a|^n}=|a|<1 \implies |a|^n \to 0$$

then since

$$-|a|^n\le a^n\le |a|^n$$

by squeeze theorem we conclude that

$$a^n \to 0$$

Answer 7

For fun:

1) $a_n= |a|^n$ is a strictly decreasing sequence for $ 0 <|a| \lt 1$.

2) $a_n$ is bounded below by $0$.

3) Hence convergent.

4)Recursively defined by:

$a_{n+1} =|a|a_n$, with $a_1=|a|$.

5) $\lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty}(|a|a_n)$.

With $\lim_{n \rightarrow \infty }a_{n+1} =$

$ \lim_{n \rightarrow \infty} a_n =:L \ge 0$,

we have:

6) $L=|a|L$ with $0 < |a| \lt 1 $.

Hence?