If $(a_n)$ is convergent, then $\liminf (a_n+b_n) = \lim a_n + \liminf b_n$

Question

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.

Answer 1

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: $$\begin{align} \lim\inf b_n&\ge \lim\inf(a_n+b_n)-\lim a_n\\ \\ \lim\inf b_n+\lim a_n&\ge \lim\inf(a_n+b_n)\end{align}$$

Answer 2

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 $\liminf_n (a_n+b_n) \le \lim_k a_{n_k}+b_{n_k} = \lim_n a_n+\liminf_n b_n$.

Answer 3

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 $\liminf$ ,

$\begin{align}\liminf b_n&\ge \liminf (a_n+b_n)+\liminf(-a_n)\\&=\liminf(a_n+b_n)-a\end{align}$

$[a_n\to a\implies -a_n\to -a]$