When do we have $\liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)$?

It's not hard to show that $$\liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)$$ for any $\{a_n\},\{b_n\}\subset{\Bbb R}$ such that the right hand side is defined (i.e. no $\infty-\infty$ or $-\infty+\infty$). Also, if both $\lim a_n$ and $\lim b_n$ exist, then we have $$ \liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n). $$

The wikipedia article about limit inferior and limit superior gives a sufficient conditions for "$=$" to hold which I don't see how to prove it:

if one of $\lim a_n$ and $\lim b_n$ exists, then we have "$=$".

Here are my questions:

• How can I show the statement above?
• Is this condition also necesarry?

Assume for example $\lim a_n=a$. To show $$ \liminf_{n\to\infty}(a_n+b_n)=a+\liminf_{n\to\infty}(b_n), $$ it suffices to show that $$ \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$ I think somehow I would need to use $\liminf_{n\to\infty}(a_n)=\limsup_{n\to\infty}(a_n)=a$. But I don't see how this works.

Solution 1:

Equality: I will use the most convenient characterization of $\liminf x_n$ as the least limit in $[-\infty,+\infty]$ of all converging subsequences of $x_n$ (including the infinity cases, where I would say, tend, rather than converge).

Take a subsequence $b_{n_k}$ of $b_n$ which tends to $\liminf b_n$. Then $$\lim_k a_{n_k}+b_{n_k}= \lim_k a_{n_k}+\lim_k b_{n_k}=\lim a_n +\liminf b_n.$$ So $\liminf (a_n+b_n)\leq \lim a_n +\liminf b_n$.

Now take a subsequence $a_{n_k}+b_{n_k}$ of $a_n+b_n$ which tends to $\liminf (a_n+b_n)$. Then $$\lim_k b_{n_k}=\lim_k (a_{n_k}+b_{n_k})-a_{n_k}=\liminf (a_n+b_n)-\lim a_n.$$ So $\liminf b_n\leq \liminf (a_n+b_n)-\lim a_n$.

This proves the desired equality.

Non equivalence: For instance $$ \liminf (-1)^n+(-1)^n=-2=(-1)+(-1)=\liminf (-1)^n+\liminf (-1)^n. $$