Question about the proof of $\liminf_{n\rightarrow\infty}(a_{n})+\liminf_{n\rightarrow\infty}(b_{n})\leq\liminf_{n\rightarrow\infty}(a_{n}+b_{n})$ I know there are multiple posts in this site about this proof but at this point I have read them all and am still struggling with a certain point.
Let $\left\{ a_{n}\right\} _{n\in\mathbb{N}}$ and $\left\{ b_{n}\right\} _{n\in\mathbb{N}}$ be bound real sequences. Then we have $$\liminf_{n\rightarrow\infty}\left(a_{n}\right)+\liminf_{n\rightarrow\infty}\left(b_{n}\right)\leq\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)$$
I am trying to understand the proof that uses the fact that both sequences have a common subsequence $\left\{ n_{k}\right\} _{k\in\mathbb{N}}$ (Whose existence I can prove) such that both $a_{n_{k}}\xrightarrow[k\rightarrow\infty]{}K$ and $b_{n_{k}}\xrightarrow[k\rightarrow\infty]{}L$. It is  clear to me that $$\liminf_{n\rightarrow\infty}\left(a_{n}\right)+\liminf_{n\rightarrow\infty}\left(b_{n}\right)\leq K+L$$ But why would it be true that $$\liminf_{n\rightarrow\infty}\left(a_{n}+b_{n}\right)=K+L$$
 A: Equality need not be true.

For example, let $a_n=(-1)^{n}$, and let $b_n=(-1)^{n+1}$.

Inequality is the best you can guarantee.

Note: For the inequality, there's no need for the subsequences approaching the respective $\lim \inf$ values to be common. In fact, if there were common subsequences approaching the respective $\lim \inf$ values, then you would get equality.

So we've already shown that equality need not hold.

To prove the inequality, we can argue as follows . . .

Suppose the sequences $(a_n)$ and $(b_n)$ are bounded.

The goals is to show
$$\lim\inf\,(a_n + b_n) \ge \lim \inf\,a_n + \lim\inf\,b_n$$

Let $K=\lim\inf\,a_n$, and let $L=\lim\inf\,b_n$.

Let $\epsilon > 0$.

Since $K=\lim\inf\,a_n$, there are at most finitely many terms of the sequence $(a_n)$ which are less than $K-\epsilon$.

Since $L=\lim\inf\,b_n$, there are at most finitely many terms of the sequence $(b_n)$ which are less than $L-\epsilon$.

Hence, at most finitely many terms of the sequence $(a_n+b_n)$ are less than $K+L-2\epsilon$.

Since $\epsilon$ can be made arbitrarily small, it follows that
$$\lim\inf\,(a_n+b_n)\ge K+L = \lim\inf\,a_n+\lim\inf\,b_n$$
as was to be shown.
A: You should start your proof differently, and first take a subsequence such that $(a_{n_k}+b_{n_k})_k$ converges to $\liminf_{n}(a_n+b_n)$. Then you take convergent subsequences of $a_{n_k}$ and $b_{n_k}$. Can you continue from there?
