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. 
 A: 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
$$
\liminf (-1)^n+(-1)^n=-2=(-1)+(-1)=\liminf (-1)^n+\liminf (-1)^n.
$$
A: Simply 
$$\liminf_{n\to\infty} (b_n)=
\liminf_{n\to\infty}(a_n-a+b_n)=\liminf_{n\to\infty} (a_n+b_n)-a
$$
Note that in the first equality $|a_n-a|$ can be made less than $\varepsilon$ for sufficiently 
large $n$   $\clubsuit$
