Construct a sequence \(\{a_n\}\) such that \(a_n > 0\) for all \(n \in \mathbb{N}\), \(\lim\limits_{n \rightarrow \infty} a_n = 0\), but \(\lim\limits_{n \rightarrow \infty} \frac{1}{a_n} = \infty\).

Definitions of sequences and convergence: here, Definitions of subsequences: here.


Solution

Consider the sequence

$$ \begin{align*} a_n = \frac{1}{n} \end{align*} $$

Then, for any \(\epsilon > 0\), choose \(N \in \mathbb{N}\) such that \(N = \lceil \frac{1}{\epsilon} \rceil + 1\). Then, for all \(n \geq N\)

$$ \begin{align*} n &\geq N \\ n &> \frac{1}{\epsilon} \\ \frac{1}{n} &< \epsilon \\ \left| \frac{1}{n} - 0 \right| &< \epsilon \quad \text{($n$ is a natural number)} \end{align*} $$

By definition of convergence, then \(\lim\limits_{n \rightarrow \infty} a_n = 0\). Next, consider the sequence

$$ \begin{align*} b_n = \frac{1}{\frac{1}{a_n}} = \frac{1}{\frac{1}{n}} = n \end{align*} $$

We want to show that

$$ \begin{align*} \lim\limits_{n \rightarrow \infty} b_n = \infty \end{align*} $$

To prove this, we must show that for every real number \(M \in \mathbb{R}\), there exists a natural number \(N\) such that when \(n \geq N\)

$$ \begin{align*} b_n = n > M \end{align*} $$

So choose \(N = \lceil M \rceil + 1\). Then, \(N > M\) and for any \(n \geq N\)

$$ \begin{align*} b_n = n &\geq N > M \end{align*} $$

Therefore, by the definition, we must have \(\lim\limits_{n \rightarrow \infty} b_n = \infty\). \(\blacksquare\)


References