Lim sup and lim inf definition

Question

$a_{n}=(-1)^n$ 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?

Answer 1

Yes, $$\liminf_{n\to\infty} a_n = 1.$$ The $\liminf$ is the value of the smallest limit point of the sequence. You have $$\limsup_{n\to\infty} a_n = +\infty.$$ since a subsequence of the $a_n$ increases without bound.