### Infinite series: Introduction

### Lim inf and lim sup

For a sequence \((a_n)_{n\in\mathbb N}\) we can construct two new sequences in the following way: \[\begin{aligned} s_N &= \inf\{ a_n : n > N \} \\ t_N &= \sup\{ a_n : n > N \}\text.\end{aligned}\] This means that \[\begin{aligned} s_1 &= \inf\{a_2, a_3,a_4, \dots\} \\ s_2 &= \inf\{a_3, a_4, a_5,\dots\} \\ s_3 &= \inf\{a_4, a_5, a_6, \dots\}\text,\end{aligned}\] etc. The set of which we take the infimum decreases as \(N\) increases as we remove the first element in each step. The infimum therefore increases if \(N\) increases. The sequence \((s_N)_{N\in\mathbb N}\) therefore is increasing and it follows that \(\lim_{N\to\infty} s_N\) exists. (The limit can be \(\infty\)). Similarly \((t_N)_{N\in\mathbb N}\) is a decreasing sequence and \(\lim_{N\to\infty} t_N\) exists as well. These two limits are called the lim inf and the lim sup of a sequence.

Let \((a_n)_{n\in\mathbb N}\) be a sequence of real numbers. We define \[ \limsup_{n\to\infty} a_n = \lim_{N\to\infty} \sup \{a_n : n > N\} \] and \[ \liminf_{n\to\infty} a_n = \lim_{N\to\infty} \inf \{a_n : n > N\} \]

Consider the sequence defined by \(a_n=(-1)^n\), or \[ -1,1,-1,1,-1,1,-1,1,\dots \] The set \(\{a_n : n > N\}\) equals \(\{-1,1\}\) for all \(N\), so \(\liminf_{n\to\infty} a_n = -1\) and \(\limsup_{n\to\infty} a_n = 1\) .

For \(b_n = \tfrac{1}{n}\) we see that \(\{b_n : n > N\} = \{\tfrac{1}{N+1}, \tfrac{1}{N+2},\tfrac{1}{N+3},\dots\}\) so \(\sup\{b_n : n > N\} = \tfrac{1}{N+1}\) and \(\inf \{b_n : n > N\} = 0\). The limits therefore both equal \(0\).

In the last example we saw that \(\lim b_n = \liminf b_n = \limsup b_n\). The following theorem says that this true in a more general setting.

The limit \(\lim_{n\to\infty} a_n\) exists if and only if \(\liminf_{n\to\infty} a_n = \limsup_{n\to\infty}a_n\) .

If this is the case, then these three limits have the same value.

For all positive sequences \((a_n)_{n=1}^\infty\) the following inequalities hold:

\[ \liminf_{n\to\infty}\ \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty}\ \sqrt[n]{a_n} \leq \limsup_{n\to\infty}\ \sqrt[n]{a_n} \leq \limsup_{n\to\infty}\ \frac{a_{n+1}}{a_n} \]