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
- Problem Statement Source: Aleph 0