I'm trying to prove below result about $\liminf$.
Let $a_n, b_n \in \mathbb R$ such that $(a_n)$ is convergent, then
$$
\liminf (a_n+b_n) = \lim a_n + \liminf b_n.
$$
- Could you have a check on my proof?
- Are there other simpler (or more direct) approaches?
My attempt: Clearly, $A :=\liminf (a_n+b_n) \ge \lim a_n + \liminf b_n$. Let's prove the reverse inequality. Let $\varphi$ be a subsequence of $\mathbb N$ such that
$$
a_{\varphi (n)} + b_{\varphi (n)} \to A, \quad n \to \infty.
$$
We have $a_{\varphi (n)} \to a :=\lim a_n$, so $b_{\varphi (n)} \to A-a$. Clearly, $A-a \ge b := \liminf b_n$. Assume the contrary that $A-a > b$. Then there is a subsequence $\psi$ of $\mathbb N$ such that $\lim_n b_{\psi (n)} < A-a$. Then
$$
A \le \lim_n (a_{\psi (n)} + b_{\psi (n)}) = a + \lim_n b_{\psi (n)}< a + (A-a) =A.
$$
Then we obtain a contradiction. This completes the proof.
Best Answer
Let's prove the reverse inequality.
$$\lim\inf b_n=\lim\inf((a_n+b_n)+(-a_n))\ge \lim\inf(a_n+b_n)+\lim\inf(-a_n)$$
Since $a_n$ is convergent, we have: $$\lim\inf(-a_n)=-\lim\sup a_n=-\lim a_n$$
Plug in and we get: $$\begin{align} \lim\inf b_n&\ge \lim\inf(a_n+b_n)-\lim a_n\\ \\ \lim\inf b_n+\lim a_n&\ge \lim\inf(a_n+b_n)\end{align}$$