How to prove $\liminf\limits_{n\rightarrow \infty}(a_n+b_n)=\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$

If $$\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$$ exists, then $$\liminf\limits_{n\rightarrow \infty}(a_n+b_n)=\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$$.

My attempt:

If either $$\liminf\limits_{n\rightarrow \infty}a_n=+\infty$$ or $$\liminf\limits_{n\rightarrow \infty}b_n=+\infty$$, then there's nothing to prove, hence we assume that $$\liminf\limits_{n\rightarrow \infty}a_n=A$$ and $$\liminf\limits_{n\rightarrow \infty}b_n=B$$, where $$A>+\infty$$ and $$B>+\infty$$. Furthermore there exists $$N_1, N_2 \in \mathbb{N}$$ such that $$a_n and $$b_n. Let $$N:=\max\{N_1,N_2\}$$, then $$\forall n\geq N:a_n+b_n

We chose $$\varepsilon$$ arbitrary, thus we have $$\liminf\limits_{n\rightarrow \infty}(a_n+b_n)= A+B$$

• I think this is false? $(a_n)_n = (0,1,0,1,0,1,\dots), (b_n)_n = (1,0,1,0,1,0,\dots)$ May 31, 2019 at 11:53
• If $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$ then lim inf $a_n$ = lim inf $b_n = -1$ but lim inf $(a_n+b_n)=0$ May 31, 2019 at 11:53
• May 31, 2019 at 11:56
• Did you mean lim inf ($a_n+b_n)\color{red}\ge$ lim inf $a_n$ + lim inf $b_n,$ or lim $(a_n+b_n)=$ lim $a_n+$ lim $b_n$? May 31, 2019 at 11:56
• in the first paragraph it says "If $\liminf\limits_{n\rightarrow \infty}a_n+\liminf\limits_{n\rightarrow \infty}b_n$ exists [...]\$ - does that change anythign? May 31, 2019 at 12:03

You've misstated what the liminf is. You should have that for $$\epsilon > 0$$ there exist $$N_1,N_2$$ so that $$n \ge N_1 \implies a_n > A - \epsilon/2$$ and $$n \ge N_2 \implies b_n > B - \epsilon/2.$$ The best you can get from this is $$n \ge \max\{N_1,N_2\} \implies A+B \le a_n + b_n + \epsilon.$$
This leads to $$A+B \le \liminf (a_n + b_n)$$. You can't get equality, as pointed out in the comments.
$$\liminf _{n\to \infty }(a_{n}+b_{n})\geq \liminf _{n\to \infty }(a_{n})+\liminf _{n\to \infty }(b_{n})$$ In the particular case that one of the sequences actually converges, say $$a_{n}\to a$$, then the inequalities above become equalities (with $$\limsup _{n\to \infty }a_{n}$$ or $$\liminf _{n\to \infty }a_{n}$$ being replaced by $$a$$).