ลองมั่ว พิสจน์ดูเอง พอจะโอเคมั้ยอ่ะครับ
$ From \ lim \inf \ x_n \ = \ A \ and \ lim \ sup \ y_n = B$
Let e>0, there exist $ N_1 \ and \ N_2$ such that if $ n\geqslant N_1 $ then $x_n \geqslant A-e/2$
and if $ n \geqslant N_2$ then $ y_n < B+e/2 $
and (*) there are infinitely many $ y_n , n \geqslant N_2$, such that $y_n \geqslant B-e/2$
From (*) construct a subsequence $ y{_n{_k}} $ define by $ y{_n{_k}} = y_n \ if \ y_n \geqslant B-e/2$
Hence if
N= max{$N_1,N_2$} and $ n\geqslant N , n_k \geqslant k \geqslant N $ then $ y{_n{_k}} + x_n > A+B - e$
So $ lim \ inf \ (y{_n{_k}} + x_n) \geqslant A+B $
Form the fact that $ lim \ inf \ (y{_n{_k}} + x_n) \leqslant lim \ sup \ (y{_n{_k}} + x_n)$
Hence $$ lim \ sup \ (y{_n{_k}} + x_n) \geqslant A+B$$
Next the expression $ lim \ sup \ (y_n + x_n) \geqslant lim \ sup \ (y{_n{_k}} + x_n) $ will be shown true
Notice that every $ n \geqslant N $ such that $ y_n < B-e/2 $ won't be any terms of $y{_n{_k}}$
So when N incerease the terms of $y{_n{_k{_m}}}$ that differ from $ lim \ sup \ y{_n{_k}}$ the least must be eliminated before the correspond term of $ y{_n{_k{_m}}}$ in the sequnce $y_n$. This imply that $ lim \ sup \ (y_n + x_n) \geqslant lim \ sup \ (y{_n{_k}} + x_n) $
Therefore $$lim \ sup \ (y_n + x_n) \geqslant A+B = \ lim \inf \ x_n \ + \ lim \ sup \ y_n$$
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