Let’s consider real sequences \((a_n)_{n \in \mathbb N}\) and \((b_n)_{n \in \mathbb N}\). We look at inequalities involving limit superior and limit inferior of those sequences. Following inequalities hold:

\[\begin{aligned}

& \liminf a_n + \liminf b_n \le \liminf (a_n+b_n)\\

& \liminf (a_n+b_n) \le \liminf a_n + \limsup b_n\\

& \liminf a_n + \limsup b_n \le \limsup (a_n+b_n)\\

& \limsup (a_n+b_n) \le \limsup a_n + \limsup b_n

\end{aligned}\] Let’s prove for example the first inequality, reminding first that \[

\liminf\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \left(\inf\limits_{m \ge n} a_m \right).\] For \(n \in \mathbb N\), we have for all \(m \ge n\) \[\inf\limits_{k \ge n} a_k + \inf\limits_{k \ge n} b_k \le a_m + b_m\] hence \[\inf\limits_{k \ge n} a_k + \inf\limits_{k \ge n} b_k \le \inf\limits_{k \ge n} \left(a_k+b_k \right)\] As the sequences \((\inf\limits_{k \ge n} a_k)_{n \in \mathbb N}\) and \((\inf\limits_{k \ge n} b_k)_{n \in \mathbb N}\) are non-increasing we get for all \(n \in \mathbb N\), \[\liminf a_n + \liminf b_n \le \inf\limits_{m \ge n} \left(a_m+b_m \right)\] which leads finally to the desired inequality \[\liminf a_n + \liminf b_n \le \liminf (a_n+b_n).\]

In fact, all inequalities above can be strict. For example, take \(a_n = (-1)^n\) and \(b_n = -(-1)^n\) we have for all \(n \in \mathbb N\): \(a_n+b_n=0\), \(\liminf a_n = \liminf b_n =-1\). Hence \(\liminf a_n + \liminf b_n = -2\) while \(\liminf a_n + b_n = 0\). We can even find sequences \((a_n)_{n \in \mathbb N}\) and \((b_n)_{n \in \mathbb N}\) such that the four inequalities are simultaneously strict. For this, consider the sequences \[

\begin{aligned}

& (a_n) = 0, 1, 2, 1, 0, 1, 2, 1, 0, \dots \text{ with } \liminf a_n=0, \, \limsup a_n=2\\

& (b_n) = 2, 1, 1, 0, 2, 1, 1, 0, 2, \dots \text{ with } \liminf b_n=0, \, \limsup b_n=2

\end{aligned}\] We have \[

\begin{aligned}

& (a_n + b_n) = 2, 2, 3, 1, 2, 2, 3, 1, 2, \dots \\

&\text{and } \liminf (a_n+b_n)=1, \, \limsup (a_n+b_n)=3

\end{aligned}\] which leads to

\[\begin{aligned}

& \liminf a_n + \liminf b_n &= 0\\

& \liminf (a_n+b_n) &= 1\\

& \liminf a_n + \limsup b_n &= 2\\

& \limsup (a_n+b_n) &= 3\\

& \limsup a_n + \limsup b_n &= 4

\end{aligned}\]

Other interesting topic is that the inequality \[\liminf a_n < \limsup b_n\] can also hold while for all \(n \in \mathbb N\) \(a_n > b_n\). This is the case for the sequences \[

\begin{aligned}

& a_n = \begin{cases}

2 & \text{if } n \text{ is odd}\\

0 & \text{if } n \text{ is even}

\end{cases}\\

& b_n = \begin{cases}

1 & \text{if } n \text{ is odd}\\

-1 & \text{if } n \text{ is even}

\end{cases}

\end{aligned}\]