Why is $\frac {1}{n^2-1} \le \frac {2}{n^2} $?

Question

I don't know how to prove that $\frac {1}{n^2-1} \le \frac {2}{n^2} $.

Can someone help me to understand?

Answer 1

You have, for any integer $n\geq 2$, $$ \frac{n^2-1}{n^2} = 1-\frac{1}{n^2} \geq 1-\frac{1}{4} > \frac{1}{2}\,. $$ Rearranging gives you the desired inequality: $$ \frac{2}{n^2} > \frac{1}{n^2-1}\,. $$

Answer 2

If $2 \leq n$ then $$ 2 + n^2 < 2n^2 \quad \Rightarrow \quad n^2 < 2(n^2 - 1) \quad \Rightarrow \quad \dfrac{2}{n^2} > \dfrac{1}{n^2-1} $$