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 $a_n<A+\varepsilon /2$ and $b_n<A+\varepsilon /2$. Let $N:=\max\{N_1,N_2\}$, then $$\forall n\geq N:a_n+b_n<A+\varepsilon /2 +B+ \varepsilon /2 =A+B+\varepsilon$$

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

  • 5
    $\begingroup$ I think this is false? $(a_n)_n = (0,1,0,1,0,1,\dots), (b_n)_n = (1,0,1,0,1,0,\dots)$ $\endgroup$ May 31, 2019 at 11:53
  • 1
    $\begingroup$ If $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$ then lim inf $a_n$ = lim inf $b_n = -1$ but lim inf $(a_n+b_n)=0$ $\endgroup$ May 31, 2019 at 11:53
  • $\begingroup$ See also math.stackexchange.com/questions/2107580/… $\endgroup$
    – Arnaud D.
    May 31, 2019 at 11:56
  • $\begingroup$ 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 $b_n$? $\endgroup$ May 31, 2019 at 11:56
  • $\begingroup$ 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? $\endgroup$
    – Analysis
    May 31, 2019 at 12:03

2 Answers 2


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.


This might be useful: Calculus - Proofs of some basic limit rules.
The R.T.P. is a basic property for limits known as "Sum Rule for Limits", But the example above isn't a normal limit, it's an inferior limit. So, I think this can help: Limit superior and limit inferior - Properties.
As written in that article:-

The limit inferior satisfies superadditivity:
$$\liminf _{n\to \infty }(a_{n}+b_{n})\geq \liminf _{n\to \infty }(a_{n})+\liminf _{n\to \infty }(b_{n})$$ 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 $a$).

I hope this helps !


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