Finding the values of $z$ such that $\sum_{n=0}^{\infty}(1+\sin{n})^nz^n$ converges

Question

I'm trying to apply the nth root test to $$\sum_{n=0}^{\infty}(1+\sin{n})^nz^n$$ Hence I use that $\hat{R}=\left (\limsup |a_n|^{\frac{1}{n}}\right )^{-1}$ and get $$\hat{R}=\left (\limsup (1+\sin{n}) \right )^{-1}$$ Now as $\sin{n}$ oscillates between $-1$ and $1$ I have that $$\hat{R}=\frac{1}{2}$$ And then for $|z|<\frac{1}{2}$ we have convergence and for $|z|>\frac{1}{2}$. However in my professors solutions he states that by the nth root test we get $\hat{R}=1$ and inputting values of $z$ less than $1$ but greater than a half in wolfram alpha gives convergence. Therefore why is $\limsup (1+\sin{n})=1$?

Answer 1

Over its period $[0,2\pi]$ the sine function is greater that $\frac{1}{\sqrt{2}}$ over the interval $\left[\frac{\pi}{4},\frac{3\pi}{4}\right]$ that has relative measure $\frac{1}{4}$, so the asymptotic density of the set of integers such that $\sin n\geq\frac{1}{\sqrt{2}}$ is $\frac{1}{4}$ by the equidistribution theorem and $$\limsup\left|1+\sin n\right|\geq 1+\frac{1}{\sqrt{2}}. \tag{1} $$ By the same argument we may replace the RHS of $(1)$ with any number in the range $(1,2)$, so the radius of convergence of the original series is $\color{red}{\large\frac{1}{2}}$, not $1$. Even professors are wrong, sometimes.