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