# If $\lim\inf_{n\to\infty}(a_n+b_n)=\lim\inf_{n\to\infty}\ a_n+\lim\inf_{n\to\infty} b_n$ for any sequence$\{b_n\}$, does $\{a_n\}$ have to converge?

If $$\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\inf}\ a_n+\underset{n\to\infty}{\lim\inf}\ b_n$$ for any sequence $$\{b_n\}$$ in $$\Bbb R$$, does $$\{a_n\}$$ have to be convergent?

My attempt:

This statement is somewhat converse from the one proven here. I considered two sequences: $$x_n=-a_n$$ and $$y_n=-b_n$$. Then $$-\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\sup}(-a_n-b_n)=\underset{n\to\infty}{\lim\sup}(x_n+y_n)$$

Instead of the definition $$\underset{n\to\infty}{\lim\inf}\ x_n=\lim_{k\to\infty}\underset{n\ge k}{\inf}\ x_n,$$ I used the fact if $$\{x_n\}$$ is bounded in $$\Bbb R,\space \underset{n\to\infty}{\lim\inf}\ x_n$$, as an accumulation point of $$\{x_n\}$$, is a limit of a convergent subsequence $$\{x_{p_n}\}_n$$ of $$\{x_n\}$$, i. e., $$\lim\limits_{n\to\infty} x_{p_n}=\underset{n\to\infty}{\lim\inf}\ x_n$$.

If the corresponding sequence $$\{y_n\}$$ is defined by $$y_n:=\begin{cases}\varepsilon,& n\in\{p_n\}\\0, &n\notin\{p_n\}\end{cases},$$ let $$\varepsilon=\underset{n\to\infty}{\lim\sup}\ x_n-\underset{n\to\infty}{\lim\inf}\ x_n$$. Then $$\color{blue}{-\underset{n\to\infty}{\lim\inf}(a_n+b_n)\ge-\underset{n\to\infty}{\lim\inf}\ a_n (*)}$$.

Since $$x_{p_n}\overset{n\to\infty}{\to}\underset{n\to\infty}{\lim\inf}$$, $$(\forall\varepsilon>0)(\exists n_\varepsilon\in\{p_n\})(n\in\{p_n\}\land n>n_\varepsilon)\implies\underbrace{(|x_n-\underset{n\to\infty}{\lim\inf}\ x_n|<\varepsilon)}_{\implies x_n<\underset{n\to\infty}{\lim\inf}\ x_n+\varepsilon}$$

For $$\varepsilon=\underset{n\to\infty}{\lim\sup}\ x_n-\underset{n\to\infty}{\lim\inf}\ x_n$$, we then obtain $$x_n<\underset{n\to\infty}{\lim\sup}\ x_n\tag 1$$.

On the other hand, $$\forall n\in\Bbb N$$ sufficiently large, it holds $$x_n<\underset{n\to\infty}{\lim\sup}\ x_n\tag 2+\varepsilon,$$ so $$(1)\land(2)\implies x_n+y_n<\underset{n\to\infty}{\lim\sup} x_n+\varepsilon$$, and it should hold $$\underset{n\to\infty}{\lim\sup}(x_n+y_n)\le\underset{n\to\infty}{\lim\sup} x_n$$, but I'm not sure how to precisely justify it. This is equivalent to $$\color{blue}{-\underset{n\to\infty}{\lim\inf}(a_n+b_n)\le-\underset{n\to\infty}{\lim\inf}\ a_n(**)}$$.

Finally, $$(*)\land(**)\implies\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\inf}\ a_n\implies\underset{n\to\infty}{\lim\sup}\ y_n=\varepsilon=0\implies\{x_n\}\text{ is convergent}$$.

May I ask if my arguments are valid and how to improve my proof if it makes any sense?

I think you are on the right track, but it becomes a bit confusing because $$\epsilon$$ is used both as the difference between $$\limsup x_n$$ and $$\liminf x_n$$ (which even might be infinite) and also as an arbitrary positive value.
I would argue as follows: Let $$(a_n)$$ be a sequence and assume that $$-\infty \le I = \liminf_{n \to \infty} a_n < S = \limsup_{n\to \infty} a_n \le \infty\, .$$
Let $$c$$ be a real number with $$I < c < S$$, and $$(a_{n_k})$$ a subsequence of $$(a_n)$$ with $$a_{n_k} \to S$$. Then define the sequence $$(b_n)$$ as $$b_n = \begin{cases} 0 & \text{ if } n \in \{ n_1, n_2, n_3,\ldots \} \\ c - a_n & \text{ otherwise.} \end{cases}$$ Then $$a_{n_k} + b_{n_k} = a_{n_k} > c$$ for all sufficiently large $$k$$, and $$b_n \ge 0$$ for all sufficiently large $$n$$. It follows that $$I < c \le \liminf_{n \to \infty} (a_n + b_n) = \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n = I + 0 = I$$ which is impossible. It follows that necessarily $$I=S$$, which means that the sequence $$(a_n)$$ is convergent.