How to prove that $ left ( frac {(- 1) ^ n} { sqrt {n}} right) _ {n geq 1} $ Is it a null sequence?

**My attempt for induction:**

If I prove that the denominator grows faster than the numerator, I can conclude that it is a null sequence, right? So I have to prove that $ sqrt {n} geq (-1) ^ n $ for $ forall n in mathbb {N}: n geq 1 $

Base case with $ n_0 = 1 $: $ sqrt {1} = 1 geq (-1) ^ 1 = -1 checkmark $

Induction hypothesis: $ exists n in mathbb {N}: sqrt {n} geq (-1) ^ n $

Induction claim: $ Longrightarrow sqrt {n + 1} geq (-1) ^ {n + 1} $

Inductive step: $$ begin {gather} sqrt {n + 1} geq (-1) ^ {n + 1} | cdot (-1) \ – sqrt {n + 1} leq (-1) ^ {n + 2} end {meets} $$ This is true, since $ – sqrt {n + 1} $ can not be $ geq -1 $.

Is that a valid test?