Proof Verification and Help: Proving $b^n$ converges to $0$

Question

Claim: $b^n \to 0$ as $n \to \infty$ for $0<b<1$

Proof: Let $ \epsilon \gt0$ be given.

Rough work:

$|b^n| = b^n \lt \epsilon$ $ \implies n\gt\frac{\ln\epsilon}{\ln b}$

Set $N = \frac{\ln\epsilon}{\ln b}$

We want to show that $ \forall n\gt N, |b^n - 0|\lt\epsilon$

Then:

I'm confused as to how to approach further than this and show that $b^n <\epsilon$

Can anyone please help in how to proceed after this and conclude my proof?

Thank you.

Answer 1

You want to order things slightly differently.

Fix $\epsilon>0$. Choose $N>\dfrac{\ln\epsilon}{\ln b}$. Then whenever $n>N$, $b^n<b^N=\epsilon$ (note that $b^n<b^N$ since $b\in(0,1)$).

Since this works for every $\epsilon>0$, $b^n\to 0$ as $n\to\infty$.

Answer 2

Since $0<b<1$, write $$ \frac{1}{b}=1+t $$ with $t>0$. Then $$ \frac{1}{b^n}=(1+t)^n\ge 1+nt $$ by Bernoulli’s inequality. This implies $$ b^n\le\frac{1}{1+nt} $$ Since $$ \frac{1}{1+nt}<\varepsilon $$ is equivalent to $$ n>\frac{\varepsilon^{-1}-1}{t} $$ we can conclude.