$a_{n}=(-1)^n$ if n is even,$ a_{n}=\sqrt{n}$ if n is odd
infimum is 1
supremum is ∞ (if we allow ∞ as supremum. If we don't allow then we say supremum does not exist) because √n increases to infinity
for any m, the set an where n>=m has infimum 1 and supremum ∞, so the limit inferior is 1 and limit superior is ∞
Is it right to my procedure?
Is there a way to make the result more formal?