I'm trying to prove, that $\limsup_{n\to\infty}|\frac{a_{n+1}}{a_n}|<1$ implies $\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}$ and I'm not sure whether this proof is correct:
Considering the rules for products of the limit superior, we have that
$\limsup_{n\to\infty}|a_n|\leq\limsup_{n\to\infty}|\frac{a_n}{a_{n-1}}|\cdot\dots\cdot\limsup_{n\to\infty|\frac{a_1}{a_0}|}\cdot\limsup_{n\to\infty}|a_0|<1^n\cdot\limsup_{n\to\infty}|a_0|$.
Thus, by taking the n-th root, this would yield
$\limsup_{n\to\infty}|a_n|^\frac{1}{n}<\limsup_{n\to\infty}|a_0|^\frac{1}{n}=1$
Am I allowed to do this last step of taking the roots?
I don't think the proof works since I am not aware of infinite limsup rules and you cannot take the square root inside, and finally it is not less than 1.
However, the assumption implies that there exists $q<1$ such that $|a_{n+1}/a_n|\leq q$. Assume for simplicity this is true for all $n$. An easy induction yields $|a_n|\leq q^n$ for all $n$. But this implies $\sqrt[n]{a_n}\leq q$ and therefore the conclusion follows.