What is the epsilon proof that $x^n \rightarrow 0$ as $n \rightarrow \infty$ provided $|x| < 1 $? I only know it's true because I know the geometric series converges, which implies its terms must tend to 0, but never seen an epsilon proof of this simple fact.


Let $|x|=\dfrac{1}{1+t}$. Since $|x|\lt 1$, we have $t\gt 0$.

By the Binomial Theorem, $(1+t)^n \ge 1+nt\gt nt$. Now it is easy to find $N$ such that if $n \gt N$, then $\dfrac{1}{nt}\lt \epsilon$.

  • $\begingroup$ This may be a dumb question, but is there a broad theorem that says for any $x \in \mathbb{R}$, we can assume $x$ takes any form (in particular $x=\frac{1}{1+t}$ for $t \in \mathbb{R}, t \geq 0$)? $\endgroup$ – alwaysiamcaesar Apr 30 at 6:02

Let us assume the limit of the sequence $\{x_n\}$ is $L$. Here $x_n=x^n$. We can easily show that limit exists by using Ratio Test (given $|x|<1$)

Since $L$ is the limit, we can always find an $N \in \mathbb{N}$ for every $\epsilon > 0\,$ s.t.
$$|x_n-L|<\epsilon \qquad \forall n\ge N$$

$\Rightarrow$$L-\epsilon <x^n<L+\epsilon$ and $L-\epsilon <x^{n+1}<L+\epsilon$

Also the above equaion imlies $(L-\epsilon)x <x^{n+1}<(L+\epsilon)x$

$\Rightarrow$ $L-\epsilon < (L+\epsilon)x$

$\Rightarrow$ $L<\epsilon \frac{1+x}{1-x}$

Now since $\epsilon$ can be arbitrarily small, you can easily show that $L=0$


You don't need the binomial theorem. $(1+t)^n \ge 1+nt$ is Bernoulli's inequality, which is easily proved by induction on $n$. This proof is suitable for an introductory algebra class, once induction has been presented.

I first saw this proof in "What is Mathematics?" by Courant and Robbins - a wonderful book for learning math.

  • 1
    $\begingroup$ Shouldn't that rather be a comment to André's answer? $\endgroup$ – t.b. Sep 16 '12 at 4:01
  • $\begingroup$ Where's the answer? $\endgroup$ – Ravi Upadhyay Sep 16 '12 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.