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?
Thank you in advance!
 A: 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.
