Prove that \(\lim\big(\frac{1}{3n^3}\big)= 0\).

For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the “show the limit” template, see this.


Problem Discussion

We want to prove that \(\lim\big(\frac{1}{3n^3}\big) = 0\). To do so we need to find \(N \in \mathbb{N}\) such that for any \(\epsilon > 0\), the tail of the sequence or the terms of the sequence end up in

$$ \begin{align*} \frac{1}{3n^3} - 0 \in (-\epsilon, \epsilon) \quad \text{whenever $n \geq N$.} \end{align*} $$

In other words

$$ \begin{align*} \big\lvert \frac{1}{3n^3} - 0 \big\rvert < \epsilon \quad \text{whenever $n \geq N$.} \end{align*} $$

Solving for \(n\)

$$ \begin{align*} \big\lvert \frac{1}{3n^3} - 0 \big\rvert &< \epsilon \\ \big\lvert \frac{1}{3n^3} \big\rvert &< \epsilon \\ \frac{1}{3n^3} &< \epsilon \quad \text{($n$ is positive)} \\ 3n^3 &> \frac{1}{\epsilon}. \\ n^3 &< \frac{1}{3\epsilon} \\ \\ n &> \frac{1}{\sqrt[3]{3\epsilon}}. \\ \end{align*} $$

From this we see that we want \(n\) to be greater than \(1/\epsilon\) in order to make the inequality works.


Formal Proof

Let \(\epsilon > 0\) be arbitrary. Choose an integer \(N\) satisfying \(N > \frac{1}{\epsilon}\). We now verify that this choice is appropriate. Let \(n \geq N\). Then,

$$ \begin{align*} \big\lvert \frac{n+1}{n} - 1 \big\rvert &= \big\lvert 1 + \frac{1}{n} - 1 \big\rvert \\ &= \big\lvert \frac{1}{n} \big\rvert \\ &= \frac{1}{n} \quad \text{($n$ is positive)} \\ &\leq \frac{1}{N} \quad \text{(since $n \geq N$)} \\ &< \epsilon. \end{align*} $$

as desired. \(\blacksquare\)


References