Why is it that these two lim inf are the same? generalization of borel cantelli lemma

Question

Why is it that $\liminf_{n \rightarrow \infty} \frac{E(\eta_n^2)}{E^2(\eta_n)} =1$ is the same as saying that

$\liminf_{n\rightarrow\infty} \frac{\sigma^2(\eta_n)}{E^2(\eta_n)} =0$

I am looking at this proof, and I'm confused about this implication, can someone please explain in words what these two limits mean and why they are the same thing.

Answer 1

$\frac {\sigma^{2}(\eta_n)} {(E\eta_n)^{2}}=\frac {E\eta_n^{2}} {(E\eta_n)^{2}}-1$ so the statement is obvious.