Prove $\text{lim} \ b^n=0$ for $0 < \vert b \vert< 1$

Question

My work starts with a supposition of $N$, so that for $n > N$ we have $\vert b \vert ^n < \epsilon$.

Since $0 < \vert b \vert < 1$, we see the logarithm with base $\vert b \vert$ is a decrescent function meaning it will invert the inequality once taken. $$\vert b \vert ^n < \epsilon $$ $$n > \text{log}_{\vert b \vert}\epsilon $$

Done the scrapwork, lets start the formal proof. Let $\epsilon > 0$. Let $N = \text{log}_{\vert b \vert}\epsilon$. If $n > N$ and $0 < \vert b \vert< 1$, then $$\vert b \vert ^n < \vert b \vert ^{\text{log}_{\vert b \vert}\epsilon} = \epsilon$$ Hence by definition, $\text{lim} \ b^n=0$

Have I done everything correctly? I am using in an implicit manner that $\vert b^n \vert = \vert b \vert ^n$, is everything fine with this? (Something is really pinching me up!)

Answer 1

You can make janmarqz's result more explicit.

Since $|b^n| = |b|^n$, we can assume that $0 < b < 1$.

Then $b = \dfrac1{1+a}$ with $a > 0$, so that, from Bernoulli's inequality, $(1+a)^n \ge 1+an $.

Therefore, since $a = \dfrac1{b}-1$,

$\begin{array}\\ b^n &=\dfrac1{(1+a)^n}\\ &\le \dfrac1{1+na}\\ &\lt \dfrac1{na}\\ &=\dfrac1{n(\frac1{b}-1)}\\ \end{array} $

To make $b^n < \epsilon$, it is enough if $\epsilon > \dfrac1{n(\frac1{b}-1)}$ or $n > \dfrac1{\epsilon(\frac1{b}-1)}$.

This is, of course, far worse than using the log, but it is completely elementary.

This is also not original with me. I first saw this in "What is Mathematics?" by Courant and Robbins, which I highly recommend and is available in paperback for less than $20.

Answer 2

If $0<|b|<1$ then $|b|=\frac{1}{1+A}$ for some $A>0$. But $(1+A)^n\ge 1+nA$ for each $n\in \Bbb N$, so you have $$|b|^n=\frac{1}{(1+A)^n}\le\frac{1}{1+nA}$$ Now taking $n$ large enough you are going to reach $\frac{1}{1+nA}<\varepsilon$, for each $\varepsilon>0$. Hence $|b|^n$ too.

Answer 3

My answer can be over killing, but it has different view point.

If $0<|b|<1$, note that

$\sum_{n=0}^{\infty}|b|^n = \frac{1}{1-|b|} <\infty$.

Thus, $\lim_{n\rightarrow \infty}|b|^n=0$ so

$|\lim_{n\rightarrow \infty} b^n|\leq |\lim_{n\rightarrow \infty}|b|^n| =0$ $\implies \lim_{n\rightarrow \infty}b^n=0$.