Two simple examples of divergent sequences are \(a_n=n\) and \(a_n=(-1)^n\text{.}\) We know the first diverges because it is not bounded, and we proved from the definition in Example 3.2.6 that the second diverges. But there is a difference. Even though \(a_n=n\) does not approach any real number as \(n\) gets large, we can say that it gets larger and larger, or that it approaches infinity. So we define a new type of limit.
Definition3.4.1.
Let \((a_n)\) be a sequence. We say that \((a_n)\) tends to infinity (or goes to infinity), and write
Just like in our previous discussion of limits, proving that a sequence goes to \(\infty\) or \(-\infty\) boils down to finding an \(n_0\) that “works”. This time it will need to work for the given \(M\text{,}\) meaning that we need to have \(a_m\gt M\) (or \(\lt M\)) when \(n\geq n_0\text{.}\)
Example3.4.2.
If \(a_n=n\text{,}\) then \(\displaystyle \lim_{n\rightarrow \infty} a_n=\infty\text{,}\) and if \(a_n=-n\text{,}\) then \(\displaystyle \lim_{n\rightarrow \infty} a_n=-\infty\text{.}\) To prove this for \(a_n=n\text{,}\) we use the Archimedean principle to find \(n_0\geq M\text{.}\) Then \(n\geq n_0\) implies \(n\geq n_0 \geq M\text{.}\) The case \(a_n=-n\) is Exercise 3.7.6.
