Prove $\liminf(a_n+b_n) \geq \liminf(a_n)+\liminf(b_n)$ for bounded sequences $a_n$ and $b_n$.
I try to prove this indirectly.
Proof: Proof: Let $M=\liminf(a_n),\;L=\liminf(b_n)$.
Assume to the contrary that $ \liminf(a_n+b_n) < M+L $. Let $K= \liminf(a_n+b_n) $. Let $ \epsilon = \frac{M+L-K}{3} >0$. Now we know that there are finitely many $ a_n < L-\epsilon $ and that there are finitely many $ b_n < M-\epsilon. $ It follows that there are finitely many $ \{a_n+b_n\} $ such that $ a_n+b_n < L+M-2\epsilon $. But $ L+M-2\epsilon = L + M -2\frac{M+L-K}{3} =\frac{L+M+2K}{3}=K+\epsilon$. So we have that there are finitely many points of $ \{a_n+b_n\} $ such that $ a_n+b_n < K+\epsilon $. But since $ K= \liminf(a_n+b_n)$, this contradicts the fact that there are infinitely many points of $ a_n+b_n $ in $ (K-\epsilon,K+\epsilon) $.
The one thing I wonder about is if it matters when $r\neq s$ for values of $a_r<L-\epsilon$ and $b_s<M-\epsilon$. Because then you may not be able to add $a_r+b_s$.
If someone could let me know how this proof looks I would appreciate it
Best Answer
Your argument is correct but one deals with it in the following manner. Let $n_1$ be the greatest value of $n$ for which $a_{n}<L-\epsilon$ and $n_2$ be the greatest value of $n$ for which $b_n<M-\epsilon $. Let $N=\max(n_1,n_2)$. If $n> N$ then $$a_n\geq L-\epsilon, b_n\geq M-\epsilon$$ so that $$a_n+b_n\geq L+M-2\epsilon =K+\epsilon$$ and hence the values of $n$ for which $$a_n+b_n<K+\epsilon$$ are all less than or equal to $N$ and form a finite set. Done!!
This is essentially the same argument of my answer which uses less symbols. The inequality $$a_n+b_n\geq K+\epsilon$$ holds for all sufficiently large values of $n$.