If $(a_n)$ is convergent, then $\liminf (a_n+b_n) = \lim a_n + \liminf b_n$ 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.
 A: 
Clearly, $\liminf (a_n+b_n) \ge \lim a_n + \liminf b_n$,

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:
$$\begin{align}
\lim\inf b_n&\ge \lim\inf(a_n+b_n)-\lim a_n\\
\\
\lim\inf b_n+\lim a_n&\ge \lim\inf(a_n+b_n)\end{align}$$
A: If $b_{n_k} \to \liminf_n b_n$ then $a_{n_k}+b_{n_k} \to \lim_n a_n+\liminf_n b_n$ and
$\liminf_n (a_n+b_n) \le \lim_k a_{n_k}+b_{n_k} = \lim_n a_n+\liminf_n b_n$.
A: Super additivity of $\liminf$ $$\liminf(a_n+b_n)\ge \liminf(a_n)+\liminf(b_n)$$
Provided $(\infty-\infty) $ doesn't occur.

Suppose $a_n\to a$.Then $\liminf a_n =\limsup a_n=a$
Then $\begin{align}\liminf(a_n+b_n)&\ge \liminf(a_n)+\liminf(b_n)\\&=a+\liminf b_n\end{align}$

$b_n=(a_n+b_n) +(-a_n) $
Now by super additivity of $\liminf$ ,
$\begin{align}\liminf b_n&\ge \liminf (a_n+b_n)+\liminf(-a_n)\\&=\liminf(a_n+b_n)-a\end{align}$
$[a_n\to a\implies -a_n\to -a]$
