# Prove that a geometric sequence converges to 0 using Bernoulli’s inequality

Here is a good problem from my first exam in Advanced Calculus (introductory real analysis) taught by Yuri Ledyaev at Western Michigan University.

Prove that $$\lim_{n \to \infty} \frac{2^n}{3^n} = 0$$.

Proof: This proof uses Bernoulli’s inequality, which states that $$(1+\alpha)^n \geq 1 + \alpha n$$ for all $$\alpha > -1$$ and $$n \in \mathbb{N}$$. The epsilon definition of the limit of a sequence says that $$\lim_{n \to \infty} a_n = L$$ if, for any $$\epsilon > 0$$, we can pick any $$n$$ above a certain threshold and always get $$\left|~a_n – L~\right| < \epsilon$$. The purpose of most proofs of this type is to use algebraic manipulation to find that threshold, which we call $$N$$. For this limit, we need to show that for all $$n>N$$, we’ll have $$\left|~\frac{2^n}{3^n} – 0~\right|<\epsilon$$, or equivalently that $$\frac{1}{\frac{2^n}{3^n}} > \frac{1}{\epsilon}$$. Observe,

$$\begin{eqnarray} \frac{1}{\frac{2^n}{3^n}} &=& \frac{1}{\left(\frac{2}{3}\right)^n} \\[1.1ex] &=& \left(\frac{1}{\frac{2}{3}}\right)^n \\[1.1ex] &=& \left(\frac{3}{2}\right)^n \\[1.1ex] &=& \left(1 + \frac{1}{2}\right)^n \\[1.1ex] &\geq& 1 + \frac{1}{2}n \\[1.1ex] &>& 1 + \frac{1}{2}N \\[1.1ex] &>& \frac{1}{\epsilon} \end{eqnarray}$$

Thus, we have $$\frac{1}{2}N > \frac{1}{\epsilon} – 1$$, so taking $$\frac{1}{2}N > \frac{1}{\epsilon}$$ will surely suffice. It follows that we need $$N > \frac{2}{\epsilon}$$, and choosing any $$n>N$$ will ensure that $$\frac{1}{\frac{2^n}{3^n}}>\frac{1}{\epsilon}$$, i.e., $$\left| ~ \frac{2^n}{3^n} – 0 ~ \right|<\epsilon$$. $$\tag*{\blacksquare}$$ (The original version of this post had “series” instead of “sequence” in the title. The content of the post is unchanged.)

This entry was posted in Math. Bookmark the permalink.