Use the second squeeze theorem and prove $\lim _{n\to \infty } \frac{(-1)^n}{n^2}=0$

Question

Use the second squeeze theorem and prove $\lim _{n\to \infty } \frac{(-1)^n}{n^2}=0$

What is squeeze second theorem

what i know about squeeze theorem is

If the limits of the sequences $\{a_n\},\{b_n\},$ and $\{b_n\},$ are convergent with $a_n ≤ b_n ≤ c_n$ and $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$ then $\lim_{n \to \infty} b_n =L$

Answer 1

use that $$\left|\frac{(-1)^n}{n^2}\right|\le \frac{1}{n^2}$$ and this tends to zero for $n$ tends to infinity

Answer 2

$$ -\frac{1}{n^2}\le \frac{(-1)^n}{n^2}\le \frac{1}{n^2}$$