I have been computing the ratius of convergence for different power series using the the root-test. In many of them, I am left with nth-root expressions that computationally approaches 1 when n goes to infinity. But I miss a formal justification for why they approaches 1. Examples include: $$\limsup (\sin(\frac{\pi n}{6})^{\frac{1}{n}}) = 1 $$ $$\limsup (n^2+1)^{\frac{1}{n}} = 1 $$ $$\limsup (\frac{1}{n^k})^{\frac{1}{n}} = 1 $$
Is there a theorem relating nth-root to different sequences?
On
For the first one:
First of all, it is undefined as a real function for infinite indexes, say
$$n=8\implies \left(\sin\frac{4\pi}3\right)^{1/8}=\sqrt[8]{-\frac{\sqrt3}2}\ldots\ldots\text{doesn't exist}$$
so I'll assume you meant $\;\sqrt[n]{\left|\sin\frac{n\pi}6\right|}\;$ , and then
$$\left|\sin\frac{\pi n}6\right|\in\left\{\;0\,,\,\frac12\,,\,\frac{\sqrt3}2\,,\,1\,,\,\frac1{\sqrt2}\;\right\}\implies$$
$$\lim_{n\to\infty}\sqrt[n]{\left|\sin\frac{\pi n}6\right|}=\begin{cases}1&,\;\;n\neq 6\Bbb N\\{}\\0&,\;\;n=6\Bbb N\end{cases}\;\implies\;\;\text{the limit doesn't exist}$$
For the second limit: $n^2 \leq (n^2+1) \leq (2n^2)$ for $n\geq1$, so we have:
$$n^{\frac{2}{n}} \leq (n^2+1)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}n^{\frac{2}{n}}$$
I'm guessing you have as a standard result that $n^{\frac{1}{n}} \rightarrow 1$ as $n \rightarrow \infty$ and $a^{\frac{1}{n}} \rightarrow 1$ as $n \rightarrow \infty$ for $a > 0$. Then by the product (both sequences on either side of the inequality go to 1) and squeeze rule, $(n^2+1)^{\frac{1}{n}} \rightarrow 1$.
As for the final one:
$$\left(\frac{1}{n^k}\right)^{\frac{1}{n}} = n^{\frac{-k}{n}} = \left(n^{\frac{1}{n}}\right)^{-k}$$
Since $k$ is fixed, and we have $n^{\frac{1}{n}} \rightarrow 1$, by the product/quotient rule, $\left(\frac{1}{n^k}\right)^{\frac{1}{n}} \rightarrow 1$.