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.)

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