Prove that the square root and exponent of a function in a $\limsup$ equals the the square root and exponent of a $\limsup$ of the function?

Question

By what property do the following equalities hold?

\begin{align*} \frac{1}{\limsup\limits_{n \to \infty} \sqrt[2n]{\left|a_n\right|}} = \left( \frac{1}{\limsup\limits_{n \to \infty} \sqrt[n]{\left|a_n\right|}} \right)^{\frac{1}{2}} \end{align*} And \begin{align*} \frac{1}{\limsup\limits_{n \to \infty} \sqrt[n]{\left|a_n^2\right|}}. = \left( \frac{1}{\limsup\limits_{n \to \infty} \sqrt[n]{\left|a_n\right|}} \right)^2 \end{align*}

I am using these two properties as elements of a proof but I want to know why this is valid (intuitively, I can tell it is).

Answer 1

This might be a case where generalizing the result makes it easier to prove...

The two properties in the question hold because $\limsup\limits_nx_n^\alpha=\left(\limsup\limits_nx_n\right)^\alpha$ for every nonnegative sequence $(x_n)$ and every positive exponent $\alpha$.

Exercise:

For every continuous increasing function $u:\mathbb R_+\to\mathbb R_+$ and every nonnegative sequence $(x_n)$, $$\limsup\limits_nu(x_n)=u\left(\limsup\limits_nx_n\right).$$