Informally speaking, the sequence \(\sqrt{n}\) "converges to infinity".
  1. Create a rigorous definition for the statement \(\lim\limits_{n \rightarrow \infty} x_n = \infty\). Use this definition to prove \(\lim\limits_{n \rightarrow \infty} \sqrt{n} = \infty\).
  2. What does your definition in (a) say about the particular sequence \((1,0,2,0,3,0,4,0,5,0)\)

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

Solution

(a) We say that

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

If and only if for every real number \(M \geq 0\), there exists an \(N \in \mathbf{N}\) such that for all \(n \geq N\), we have

$$ \begin{align*} x_n > M \end{align*} $$

Now, we want to show that

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

To do this, let \(M \geq 0\), we want to find an \(N\) such that for all \(n \geq N\), we have

$$ \begin{align*} \sqrt{n} > M \end{align*} $$

Squaring both sides, we see that \(n\) needs to be greater than \(M^2\). Therefore, for every real number \(M > 0\), let \(N = \lceil M^2 \rceil + 1\). Then for for any \(n \geq N\), we must have \(\sqrt{n} > M\) and by definition, this implies that

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

as desired. \(\ \blacksquare\)

(b) Consider the sequence

$$ \begin{align*} \{a_n\}= (1,0,2,0,3,0,4,0,5,0) \end{align*} $$

Suppose that it converges to \(\infty\). Then by definition, every \(M \geq 0\), there exists an \(N\) such that for all \(n \geq N\)

$$ \begin{align*} a_n > M \end{align*} $$

But no matter how large \(N\) is, there will be infinitely many even values of \(n \geq N\) where \(a_n = 0\). So, the inequality \(a_n > M\) fails for those \(n\) values. Therefore, the definition is no longer is satisfied. Thus, this sequence doesn’t converge to \(\infty\).

References
Aleph 0